Sequence and Series

Given \[ T_p = a + (p-1)d = q \quad (i) \] \[ T_q = a + (q-1)d = p \quad (ii) \] From \((i)\) and \((ii)\), \[ pd - qd = q - p \] \[ (p-q)d = (q-p) \quad \Rightarrow \quad d = -1 \] From \((i)\), \[ a - q + 1 = p \quad \Rightarrow \quad a = p+q-1 \] Now, \[ T_{pq} = a + (pq-1)d \] \[ = (p+q-1) + (pq-1)(-1) \] \[ = p+q-1 - pq + 1 = p+q - pq \]

Consider an arithmetic series \[ a,\; a+d,\; a+2d,\; \ldots,\; a+(n-1)d \] Sum of equidistant terms are \[ T_1 + T_n = a + \big(a + (n-1)d\big) = 2a + (n-1)d \] \[ T_2 + T_{n-1} = (a+d) + \big(a + (n-2)d\big) = 2a + (n-1)d \] \[ T_3 + T_{n-2} = (a+2d) + \big(a + (n-3)d\big) = 2a + (n-1)d \] \[ \vdots \] \[ T_k + T_{n-k+1} = 2a + (n-1)d \quad \forall\; k \] Therefore, the sum of any two terms equidistant from the ends is constant and equals \(2a+(n-1)d\).

Let the arithmetic sequence be \[ a, \; a+d, \; a+2d, \; \ldots, \; a+(n-1)d \] Add a constant \(k\) to each term \[ (a+k), \; (a+d+k), \; (a+2d+k), \; \ldots, \; (a+(n-1)d+k) \] The new sequence still satisfies the definition of arithmetic series, since \[ [(a+(n-1)d+k) - (a+(n-2)d+k)] = d \] Similarly, subtracting a constant \(k\) gives \[ (a-k), \; (a+d-k), \; (a+2d-k), \; \ldots, \; (a+(n-1)d-k) \] Again, the new sequence satisfies the definition of arithmetic series, since \[ [(a+(n-1)d-k) - (a+(n-2)d-k)] = d \]

Let the arithmetic sequence be \[ a, \; a+d, \; a+2d, \; \ldots, \; a+(n-1)d \] Multiply each term by a constant \(k\): \[ ka, \; k(a+d), \; k(a+2d), \; \ldots, \; k(a+(n-1)d) \] The common difference becomes \[ [k(a+(r+1)d) - k(a+rd)] = kd \] which is constant, hence the new sequence is arithmetic. Similarly, dividing each term by a nonzero constant \(k\) gives \[ \frac{a}{k}, \; \frac{a+d}{k}, \; \frac{a+2d}{k}, \; \ldots, \; \frac{a+(n-1)d}{k} \] with common difference \[ \frac{(a+(r+1)d)}{k} - \frac{(a+rd)}{k} = \frac{d}{k} \] which is also constant. Thus, the result is an arithmetic sequence.

Given \[ \frac{b+c}{a} + 1 = \frac{a+b+c}{a}, \quad \frac{c+a}{b} + 1 = \frac{a+b+c}{b}, \quad \frac{a+b}{c} + 1 = \frac{a+b+c}{c} \] Hence \[ \frac{a+b+c}{a}, \; \frac{a+b+c}{b}, \; \frac{a+b+c}{c} \; \text{is an arithmetic sequence.} \] Divide through by \((a+b+c) \neq 0\): \[ \frac{1}{a}, \; \frac{1}{b}, \; \frac{1}{c} \; \text{is an arithmetic sequence.} \]

Start with the given sequence: \[ \frac{b+c-a}{a}, \; \frac{c+a-b}{b}, \; \frac{a+b-c}{c} \] Add 1 to each term. This transforms the sequence into: \[ \frac{a+b+c}{a}, \; \frac{a+b+c}{b}, \; \frac{a+b+c}{c} \] Since adding the same constant to each term of a sequence preserves the arithmetic property, this is also an arithmetic sequence. Now, divide each term by \((a+b+c)\): \[ \frac{1}{a}, \; \frac{1}{b}, \; \frac{1}{c} \] Dividing by a nonzero constant also preserves the arithmetic property. Therefore, \(\tfrac{1}{a}, \tfrac{1}{b}, \tfrac{1}{c}\) is an arithmetic sequence.

Start with the sum: \[ \frac{1}{a_1a_2} + \frac{1}{a_2a_3} + \cdots + \frac{1}{a_{n-1}a_n} \] Notice that in an arithmetic sequence, the difference between consecutive terms is \(d\), i.e., \[ a_{k+1} - a_k = d \] Multiply and divide each term by this difference: \[ = \frac{1}{d}\left(\frac{a_2-a_1}{a_1a_2} + \frac{a_3-a_2}{a_2a_3} + \cdots + \frac{a_n-a_{n-1}}{a_{n-1}a_n}\right) \] Simplify each fraction: \[ = \frac{1}{d}\left(\frac{1}{a_1} - \frac{1}{a_2} + \frac{1}{a_2} - \frac{1}{a_3} + \cdots + \frac{1}{a_{n-1}} - \frac{1}{a_n}\right) \] This is a telescoping series where intermediate terms cancel, leaving: \[ = \frac{1}{d}\left(\frac{1}{a_1} - \frac{1}{a_n}\right) \] Combine into a single fraction: \[ = \frac{a_n-a_1}{da_1a_n} \] But since \(a_n = a_1+(n-1)d\), we get \[ a_n - a_1 = (n-1)d \] Substituting this, \[ = \frac{(n-1)d}{da_1a_n} = \frac{n-1}{a_1a_n} \]\

Start with the given sum: \[ \frac{1}{\sqrt{a_1}+\sqrt{a_2}}+\frac{1}{\sqrt{a_2}+\sqrt{a_3}}+\ldots+\frac{1}{\sqrt{a_{n-1}}+\sqrt{a_n}} \] Multiply numerator and denominator of each term by the difference of the square roots. Since \(a_{k+1}-a_k = d\), \[ = \frac{1}{d}\Bigg[\frac{a_2-a_1}{\sqrt{a_1}+\sqrt{a_2}}+\frac{a_3-a_2}{\sqrt{a_2}+\sqrt{a_3}}+\ldots+\frac{a_n-a_{n-1}}{\sqrt{a_{n-1}}+\sqrt{a_n}}\Bigg] \] Simplify each fraction using \(a_{k+1}-a_k = (\sqrt{a_{k+1}}-\sqrt{a_k})(\sqrt{a_{k+1}}+\sqrt{a_k})\): \[ = \frac{1}{d}\Big[(\sqrt{a_2}-\sqrt{a_1})+(\sqrt{a_3}-\sqrt{a_2})+\ldots+(\sqrt{a_n}-\sqrt{a_{n-1}})\Big] \] This telescopes to: \[ = \frac{1}{d}(\sqrt{a_n}-\sqrt{a_1}) \] Multiply numerator and denominator by \(\sqrt{a_n}+\sqrt{a_1}\): \[ = \frac{1}{d}\frac{a_n-a_1}{\sqrt{a_n}+\sqrt{a_1}} \] Substituting \(a_n-a_1 = (n-1)d\), \[ = \frac{n-1}{\sqrt{a_1}+\sqrt{a_n}} \]

The last number in the \(n^{\text{th}}\) term is \[ T_n = 1 + (n-1)\cdot 2 = 2n-1 \] Thus, the \(n^{\text{th}}\) term is the sum of an arithmetic series with first term \(1\), last term \(2n-1\), and number of terms equal to \(2n-1\). Using the sum formula: \[ S_n = \frac{2n-1}{2}\big[2+(2n-2)\big] \] Simplify: \[ S_n = (2n-1)\cdot n \] \[ S_n = 2n^2-n \]

Notice that each bracket contains an arithmetic series. In the \(n^{\text{th}}\) bracket, the last term of this series is \[ T = n^2 \] The number of terms in this series is \(2n-1\). Therefore, its sum is \[ S_n = \frac{2n-1}{2}\Big[2n^2-(2n-2)\Big] \] Simplifying gives \[ S_n = \frac{(2n-1)(2n^2-2n+2)}{2} \]

Since the \(n\) arithmetic means form an arithmetic progression: \[ AM_k = a + k d \quad \text{for } k = 1, 2, \ldots, n \] The sum of these means is: \[ \sum_{k=1}^{n} AM_k = \sum_{k=1}^{n} (a + k d) \] Break into two parts: \[ = \sum_{k=1}^n a + \sum_{k=1}^n k d = na + d \sum_{k=1}^{n} k \] Since \[ \sum_{k=1}^{n} k = \frac{n(n+1)}{2}, \] we have \[ \sum_{k=1}^{n} AM_k = na + d \cdot \frac{n(n+1)}{2} \] Substituting \(d = \frac{b - a}{n+1}\): \[ \sum_{k=1}^{n} AM_k = na + \frac{b-a}{n+1} \cdot \frac{n(n+1)}{2} \] Simplify: \[ = na + \frac{n(b - a)}{2} \] \[ = \frac{2na + n(b - a)}{2} \] \[ = \frac{n(a+b)}{2} \] \[ = n \cdot \frac{a+b}{2} \] Thus, the sum of the \(n\) arithmetic means equals \(n\) times the arithmetic mean of \(a\) and \(b\).

Consider the sequence \[ a, \; ar, \; ar^2, \; \ldots, \; ar^{n-1}. \] The \(k^{\text{th}}\) term from the beginning is \[ T_k = ar^{k-1}. \] The \(k^{\text{th}}\) term from the end is \[ T_{n-k+1} = ar^{n-k}. \] Their product is \[ T_k \cdot T_{n-k+1} = \big(ar^{k-1}\big)\big(ar^{n-k}\big) = a^2 r^{n-1}. \] This expression is independent of \(k\), hence constant. Therefore, the product of any two terms equidistant from both ends is the same.

Let the original geometric sequence be \[ a, \; ar, \; ar^2, \; \ldots, \; ar^{n-1}. \]
Case 1: Squaring or raising to a power
Squaring gives \[ a^2, \; (ar)^2, \; (ar^2)^2, \ldots = a^2, \; a^2r^2, \; a^2r^4, \ldots \] The new common ratio is \(r^2\). More generally, raising to power \(m\) gives common ratio \(r^m\).
Case 2: Multiplying by a constant \(k\) \[ ka, \; kar, \; kar^2, \ldots \] The ratio remains \(r\).
Case 3: Dividing by a constant \(k \ne 0\)
\[ \tfrac{a}{k}, \; \tfrac{ar}{k}, \; \tfrac{ar^2}{k}, \ldots \] The ratio is again \(r\). Thus, in all cases, the new sequence remains geometric.

Let the \(n\) geometric means be \(g_1, g_2, \ldots, g_n\). By definition, \[ g_1 \cdot g_2 \cdot \ldots \cdot g_n = (ar)(ar^2)\cdots(ar^n) \] Simplify: \[ = a^n \cdot r^{1+2+\cdots+n} = a^n \cdot r^{\frac{n(n+1)}{2}} \] Substituting \( r = \left(\frac{b}{a}\right)^{\frac{1}{n+1}} \): \[ = a^n \cdot \left(\frac{b}{a}\right)^{\frac{n(n+1)}{2(n+1)}} \] \[ = a^n \cdot \left(\frac{b}{a}\right)^{\frac{n}{2}} \] \[ = (ab)^{\frac{n}{2}} \] But the geometric mean between \(a\) and \(b\) is \(\sqrt{ab}\). Therefore, the product of the \(n\) geometric means is \[ (\sqrt{ab})^n \] which proves the result.

Let \[ S = 1 + xy + x^2y^2 + x^3y^3 + \cdots \] Multiply by \(xy\): \[ xyS = xy + x^2y^2 + x^3y^3 + \cdots \] Subtract: \[ S - xyS = 1 \quad \Rightarrow \quad S = \frac{1}{1-xy} \] From the given conditions: \[ \frac{1}{1-x} = a \quad \Rightarrow \quad x = \frac{a-1}{a} \] \[ \frac{1}{1-y} = b \quad \Rightarrow \quad y = \frac{b-1}{b} \] Substituting: \[ S = \frac{1}{1 - \frac{(a-1)(b-1)}{ab}} \] Simplify: \[ S = \frac{ab}{a+b-1} \]

Grouping terms: \[ a + b + a^2 + b^2 + a^3 + b^3 + \cdots = (a + a^2 + a^3 + \cdots) + (b + b^2 + b^3 + \cdots) \] Each group is a geometric series: \[ = \frac{a}{1-a} + \frac{b}{1-b}, \quad \text{for } |a|<1, |b|<1 \]

Since \(a_1, a_2, \ldots, a_n\) is a harmonic sequence, their reciprocals form an arithmetic sequence. Let the common difference be \(d\). \[ d = \frac{\tfrac{1}{a_n} - \tfrac{1}{a_1}}{n-1} = \frac{a_1 - a_n}{a_1 a_n (n-1)} \] For each pair of consecutive terms: \[ \frac{1}{a_2} - \frac{1}{a_1} = d, \quad \frac{1}{a_3} - \frac{1}{a_2} = d, \quad \ldots, \quad \frac{1}{a_n} - \frac{1}{a_{n-1}} = d \] Rearranging: \[ a_1 - a_2 = d a_1 a_2, \quad a_2 - a_3 = d a_2 a_3, \quad \ldots, \quad a_{n-1} - a_n = d a_{n-1}a_n \] Summing these equations: \[ (a_1 - a_n) = d \bigl(a_1a_2 + a_2a_3 + \cdots + a_{n-1}a_n\bigr) \] Substituting the value of \(d\): \[ a_1 - a_n = \frac{a_1 - a_n}{a_1 a_n (n-1)} \cdot (a_1a_2 + a_2a_3 + \cdots + a_{n-1}a_n) \] Cancelling \((a_1 - a_n)\) (assuming it is non-zero): \[ (n-1)a_1a_n = a_1a_2 + a_2a_3 + \cdots + a_{n-1}a_n \] which is the required result.

Let \[ A = \sum_{r=1}^{b-1} \left\lfloor\frac{ar}{b}\right\rfloor \] Using , we can write \[ A = \sum_{r=1}^{b-1} \left\lfloor \frac{a(b-r)}{b} \right\rfloor \] Simplify: \[ A = \sum_{r=1}^{b-1} \left\lfloor a - \frac{ar}{b} \right\rfloor \] Since \(a\) is integer, this becomes \[ A = \sum_{r=1}^{b-1} \left(a + \left\lfloor -\tfrac{ar}{b} \right\rfloor\right) \] Adding the two expressions for \(A\): \[ 2A = \sum_{r=1}^{b-1} \left(a + \left\lfloor -\tfrac{ar}{b} \right\rfloor + \left\lfloor \tfrac{ar}{b} \right\rfloor\right) \] By the property of floor function, \(\left\lfloor -x \right\rfloor + \left\lfloor x \right\rfloor = -1\) (when \(x\) is not integer). Since \(\tfrac{ar}{b}\) is not integer (as \(a,b\) are coprime), we have \[ 2A = \sum_{r=1}^{b-1} (a-1) \] Therefore \[ 2A = (b-1)(a-1) \quad \Rightarrow \quad A = \frac{(b-1)(a-1)}{2} \]

The general term is \[ T_n = n(n+1) = n^2 + n \] So, \[ S_n = \sum (n^2 + n) = \sum n^2 + \sum n \] Using : \[ \sum n^2 = \frac{n(n+1)(2n+1)}{6}, \quad \sum n = \frac{n(n+1)}{2} \] Substitute to obtain the final simplified expression.

The general term is \[ T_n = (n+1)(n+2)(n+3) \] Expand: \[ T_n = n^3 + 6n^2 + 11n + 6 \] Hence, \[ S_n = \sum (n^3 + 6n^2 + 11n + 6) = \sum n^3 + 6\sum n^2 + 11\sum n + 6\sum 1 \] Use to calculate the final expression.

The general term is \[ T_n = (2n-1)(2n+1) = 4n^2 - 1 \] So, \[ S_n = \sum (4n^2 - 1) = 4 \sum n^2 - \sum 1 \] Apply to obtain the final expression.

The general term is \[ T_n = (n+1)(n+2)(n+3)(n+4)(n+5)(n+6)(n+7) \] To apply the \(V_n\) method, we define \[ V_n = (n+1)(n+2)(n+3)(n+4)(n+5)(n+6)(n+7)(n+8) \] Then \[ V_{n-1} = n(n+1)(n+2)(n+3)(n+4)(n+5)(n+6)(n+7) \] Subtracting, \[ V_n - V_{n-1} = (n+1)(n+2)\cdots(n+7)\,(n+8-n) \] \[ \Rightarrow V_n - V_{n-1} = 8T_n \] Hence, \[ T_n = \frac{V_n - V_{n-1}}{8} \] This gives a telescoping series: \[ S_n = \frac{1}{8}(V_n - V_0) \] Therefore, \[ S_n= \frac{1}{8} \Big[(n+1)(n+2)\cdots(n+8) - 1\cdot2\cdot3\cdot4\cdot5\cdot6\cdot7\cdot8\Big] \]

The general term is \[ T_n = (2n-1)(2n+1)(2n+3)(2n+5)(2n+7) \] Define \[ V_n = (2n-1)(2n+1)(2n+3)(2n+5)(2n+7)(2n+9) \] Then \[ V_{n-1} = (2n-3)(2n-1)(2n+1)(2n+3)\cdots(2n+7) \] So, \[ V_n - V_{n-1} = (2n-1)(2n+1)(2n+3)\cdots(2n+7)\,[(2n+9) - (2n-3)] \] \[ \Rightarrow V_n - V_{n-1} = 12T_n \] Hence, \[ T_n = \frac{V_n - V_{n-1}}{12} \] Summing telescopically, \[ S_n = \frac{1}{12}(V_n - V_0) \] So, \[ S_n = \frac{1}{12}\Big[(2n-1)(2n+1)(2n+3)\cdots(2n+9) - 1 \cdot 3 \cdot 5 \cdot 7 \cdot 9 \Big] \]

The general term is \[ T_n = \frac{1}{n(n+1)(n+2)(n+3)(n+4)} \] Define \[ V_n = \frac{1}{(n+1)(n+2)(n+3)(n+4)} \] Then \[ V_{n-1} = \frac{1}{n(n+1)(n+2)(n+3)} \] Subtract: \[ V_n - V_{n-1} = \frac{n-(n+4)}{n(n+1)(n+2)(n+3)(n+4)} = -4T_n \] So, \[ T_n = \frac{V_{n-1} - V_n}{4} \] The series telescopes: \[ S_n = \frac{1}{4}(V_0 - V_n) \] As \(n \to \infty\), \[ S_\infty = \frac{1}{4}V_0 = \frac{1}{4 \cdot (1\cdot2\cdot3\cdot4)} = \frac{1}{96} \]

The general term is \[ T_n = \frac{1}{(n+1)(n+2)(n+3)(n+4)} \] Define \[ V_n = \frac{1}{(n+2)(n+3)(n+4)} \] Then \[ V_{n-1} = \frac{1}{(n+1)(n+2)(n+3)} \] Subtract: \[ V_n - V_{n-1} = \frac{(n+1)-(n+4)}{(n+1)(n+2)(n+3)(n+4)} = -3T_n \] So, \[ T_n = \frac{V_{n-1} - V_n}{3} \] By telescoping, \[ S_n = \frac{1}{3}(V_0 - V_n) = \frac{1}{3}\left(\frac{1}{2\cdot3\cdot4} - \frac{1}{(n+2)(n+3)(n+4)}\right) \] As \(n \to \infty\), \[ S_\infty = \frac{1}{3}V_0 = \frac{1}{3 \cdot (2\cdot3\cdot4)} = \frac{1}{72} \]

We use the identity \[ (a+b+c+d)^2 = a^2+b^2+c^2+d^2 + 2(ab+ac+ad+bc+bd+cd) \] This shows that the cross-product sum can be extracted from the square of the sum.

Applying this to the first \(n\) natural numbers: \[ (1+2+3+\ldots+n)^2 = (1^2+2^2+3^2+\ldots+n^2) + 2S_n \]

Rearranging, \[ S_n = \frac{(1+2+3+\ldots+n)^2 - (1^2+2^2+3^2+\ldots+n^2)}{2} \]

Thus in summation form, \[ S_n = \frac{\Big(\sum n\Big)^2 - \sum n^2}{2} \]

Method 1: Separate into odd and even squares. \[ S_n = \Big[1^2 + 3^2 + 5^2 + \ldots + (2n-1)^2\Big] - \Big[2^2 + 4^2 + \ldots + (2n)^2\Big] \]

That is, \[ S_n = \sum (2n-1)^2 - \sum (2n)^2 \]

Method 2: Group as pairs. \[ S_n = (1-2)(1+2) + (3-4)(3+4) + \ldots + (2n-1)(2n+1) \] Each bracket simplifies to a negative consecutive sum, giving \[ \Rightarrow S_n = -(1+2+3+\ldots+2n) \]

Finally, \[ S_n = -2n(2n+1)/2 = -n(2n+1) \]

Method 1: Write the general term. \[ T_n = (2n-1)^2 - 3(2n)^2 = (4n^2 - 4n + 1) - 12n^2 = -8n^2 - 4n + 1 \]

Summing, \[ S_n = \sum_{i=1}^n \big[(2i-1)^2 - 3(2i)^2\big] = \sum_{i=1}^n (2i-1)^2 - 3\sum_{i=1}^n (2i)^2 \]

Method 2: Split into two series. \[ S_n = \Big(1^2+2^2+3^2+\ldots+(2n)^2\Big) - 4\Big(2^2+4^2+6^2+\ldots+(2n)^2\Big) \]

So, \[ S_n = \sum_{r=1}^{2n} r^2 - 4\sum_{r=1}^{n} (2r)^2 \] \[ \Rightarrow S_n = \sum_{r=1}^{2n} r^2 - 16\sum_{r=1}^{n} r^2 \]

Note that \[ S=1 + 2 + 4 + 7 + 11 + 16 + \ldots \quad n \text{ terms} \] and shifting by one term, \[ S=\quad\quad 1 + 2 + 4 + 7 + 11 + 16 + \ldots \quad n \text{ terms} \]

Subtracting the two gives \[ 0 = 1 + (1+2+3+4+\ldots+(n-1)) - T_n \] where \(T_n\) is the \(n\)th term.

Thus \[ T_n = 1 + \frac{n^2}{2} - \frac{n}{2} \] and hence \[ S_n = n + \frac{1}{2}\sum n^2 - \frac{1}{2}\sum n \]

Note that \[ S=1 + 3 + 6 + 10 + 15 + \ldots \quad \text{n terms} \] and shifting, \[ S=\quad\quad 1 + 3 + 6 + 10 + 15 + \ldots \quad \text{n terms} \]

Subtracting the two gives \[ 1 - T_n + 2 + 4 + \ldots + (n-1) = 0 \]

Hence \[ T_n = 1 + 2 + 3 + \ldots + n = \frac{n(n+1)}{2} \] and \[ S_n = \frac{1}{2}\sum n^2 + \frac{1}{2}\sum n \]

Let \[ S = 1 + 2x + 3x^2 + \ldots + (n-1)x^{n-2} + nx^{n-1} \] Multiplying by \(x\), \[ xS = x + 2x^2 + 3x^3 + \ldots + (n-1)x^{n-1} + nx^n \] Subtracting, \[ S(1-x) = 1 + x + x^2 + \ldots + x^{n-1} - nx^n \]

The geometric sum gives \[ S(1-x) = \frac{1 - x^n}{1-x} - nx^n \] so \[ S = \frac{1 - x^n}{(1-x)^2} - \frac{nx^n}{1-x} \]

Multiplying by \(x\), \[ xS = x^2 + 3x^3 + 5x^4 + \ldots + (2n-1)x^{n+1} \] Subtracting, \[ S(1-x) = x + 2x^2 + 2x^3 + \ldots + 2x^n - (2n-1)x^{n+1} \]

So \[ S(1-x) = x + \frac{2x^2(1-x^{n-1})}{1-x} - (2n-1)x^{n+1} \] \[ \Rightarrow S_n = \frac{x}{1-x} + \frac{2x^2(1-x^{n-1})}{(1-x)^2} - \frac{(2n-1)x^{n+1}}{1-x} \]

For \(n\to\infty\) (with \(|x|<1\)), \[ S_\infty = \frac{x}{1-x} + \frac{2x^2}{(1-x)^2} \]

Multiplying by \(x\), \[ xS = 1^2x^2 + 2^2x^3 + 3^2x^4 + \cdots + n^2x^{n+1} \] Subtracting, \[ S(1-x) = S' - n^2x^{n+1} \quad (i) \] where \[ S' = x + 3x^2 + 5x^3 + \cdots + (2n-1)x^n + n^2x^n \]

Now, \[ xS' = x^2 + 3x^3 + 5x^4 + \cdots + (2n-1)x^{n+1} \] Subtracting, \[ S'(1-x) = x + 2x^2 + 2x^3 + \cdots + 2x^n - (2n-1)x^{n+1} \] \[ \Rightarrow S' = \frac{x}{1-x} + \frac{2x^2(1-x^n)}{(1-x)^2} - \frac{(2n-1)x^{n+1}}{1-x} \]

Finally, substitute \(S'\) into equation (i) to obtain the expression for \(S\).

Using AM \(\geq\) GM: \[ \frac{x+\tfrac{1}{x}}{2} \geq 1 \quad\Rightarrow\quad x+\frac{1}{x} \geq 2 \] Hence, the range is \[ [2, \infty) \]

For \(x<0\), write \[ \frac{x+1}{x} = 1 + \frac{1}{x} \] Since \(x<0\), we have \(\frac{1}{x}<0\), so \[ \frac{x+1}{x} \leq 1 \] and as \(x \to 0^-\), \(\frac{x+1}{x} \to -\infty\). Thus, the range is \[ (-\infty, 1) \]

By AM \(\geq\) GM, \[ \frac{a^2+b^2}{2} \geq ab \quad (i) \] \[ \frac{b^2+c^2}{2} \geq bc \quad (ii) \] \[ \frac{c^2+a^2}{2} \geq ca \quad (iii) \] Adding \((i)+(ii)+(iii)\) gives \[ a^2+b^2+c^2 \geq ab+bc+ca \]

First, note \[ a^4+b^4+c^4 \geq a^2b^2+b^2c^2+c^2a^2 \] Now, by AM–GM: \[ \frac{a^2b^2+a^2c^2}{2} \geq a^2bc \quad (i) \] \[ \frac{a^2b^2+b^2c^2}{2} \geq b^2ca \quad (ii) \] \[ \frac{b^2c^2+c^2a^2}{2} \geq c^2ab \quad (iii) \] Adding gives \[ a^2b^2+b^2c^2+c^2a^2 \geq a^2bc+b^2ca+c^2ab = abc(a+b+c) \] Hence, \[ a^4+b^4+c^4 \geq abc(a+b+c) \]

By AM \(\geq\) GM, \[ \frac{\tfrac{a}{c}+\tfrac{b}{c}+\tfrac{b}{a}+\tfrac{c}{a}+\tfrac{c}{b}+\tfrac{a}{b}}{6} \geq 1 \] Hence, \[ \frac{a+b}{c}+\frac{b+c}{a}+\frac{c+a}{b} \geq 6 \]

Apply AM–GM to each product term. For example, \[ \frac{a}{d} \cdot \frac{d}{b} \geq 2\sqrt{\frac{a}{b}} \] Expanding all 9 terms, and applying AM–GM, gives \[ \left(\frac{a}{d}+\frac{b}{e}+\frac{c}{f}\right)\left(\frac{d}{b}+\frac{e}{c}+\frac{f}{a}\right) \geq 9 \]

By AM–GM: \[ \frac{a+b+c}{3} \geq (abc)^{1/3} \] Substituting \(a+b+c=5\), \[ \frac{5}{3} \geq (abc)^{1/3} \] Cubing, \[ abc \leq \frac{125}{27} \]

Write as: \[ \frac{a}{2}+\frac{a}{2}+\frac{b}{3}+\frac{b}{3}+\frac{b}{3}+\frac{c}{4}+\frac{c}{4}+\frac{c}{4}+\frac{c}{4} = 5 \] By AM–GM, \[ \frac{5}{9} \geq (a^2 b^3 c^4)^{1/9} \] Hence, \[ a^2 b^3 c^4 \leq \left(\frac{5}{9}\right)^9 \]

Write as: \[ \frac{3a}{2}+\frac{3a}{2}+\frac{2b}{3}+\frac{2b}{3}+\frac{2b}{3}+\frac{c}{4}+\frac{c}{4}+\frac{c}{4}+\frac{c}{4} = 5 \] By AM–GM, \[ \frac{5}{9} \geq (a^2 b^3 c^4)^{1/9} \] Hence, \[ a^2 b^3 c^4 \leq \left(\frac{5}{9}\right)^9 \]

General term: \[ T_n = n(n+1)(n+2)(n+1) \] Simplify: \[ T_n = \tfrac{1}{2}\,n(n+1)(n+2)(n+3) + \tfrac{1}{2}(n-1)n(n+1)(n+2) \] So the sum reduces to the sum of two polynomial series.