Binomial Theorem

\[ T_7 = T_{6+1} = \binom{9}{6} \cdot 3^{9-6} \cdot x^6 = \binom{9}{6} \cdot 3^3 \cdot x^6 \]

Expand the first bracket \[ (x^2 - 4x + 4)(x+7)^{13} \] Coefficient of \(x^2\) is \[ 7^{13} - 4 \binom{13}{1} 7^{12} + 4 \binom{13}{2} 7^{11} \]

\[ \text{Coeff of } x^2 = (-7)^{13} - 4 \binom{13}{1} 7^{12} + 4 \binom{13}{2} 7^{11} \] \[ = -7^{13} + 4 \binom{13}{1} 7^{12} + 4 \binom{13}{2} 7^{11} \]

General term in the binomial expansion is given by \[ T_{r+1} = \binom{51}{r} (x^2)^{51-r} \left(-\frac{1}{x}\right)^r \] \[ = \binom{51}{r} (-1)^r x^{2(51-r)} \cdot x^{-r} \] \[ = \binom{51}{r} (-1)^r x^{102 - 3r} \] We need \(102 - 3r = 15 \implies r = 29\). \[ T_{30} = \binom{51}{29} (-1)^{29} x^{15} \] Therefore, coefficient of \(x^{15}\) is \( -\binom{51}{29} \)

\[ (x+2)^6 (x+1)^6 \] Coefficient of \(x^{11}\) is \[ \binom{6}{0} \cdot x^6 \cdot \binom{6}{5} x^5 + \binom{6}{1} x^5 \cdot \binom{6}{6} x^6 \] \[ = 6x^{11} + 12x^{11} = 18x^{11} \]

\[ S = (1+x) + (1+x)^2 + \dots + (1+x)^n \] This is a GP: \[ S = \frac{(1+x)((1+x)^n - 1)}{(1+x) - 1} \] \[ = \frac{(1+x)((1+x)^n - 1)}{x} = \frac{(1+x)^{n+1} - (1+x)}{x} \] Coefficient of \(x^r\) in \(S\) comes from coeff. of \(x^{r+1}\) in numerator. \[ \Rightarrow \binom{n+1}{r+1} x^r \]

We know \[ (1+x)^n = \binom{n}{0} + \binom{n}{1}x + \binom{n}{2}x^2 + \dots + \binom{n}{n}x^n \] Differentiate both sides \[ n(1+x)^{n-1} = \binom{n}{1} + 2\binom{n}{2}x + 3\binom{n}{3}x^2 + \dots + n\binom{n}{n}x^{n-1} \] Putting \(x=1\) \[ n \cdot 2^{n-1} = \binom{n}{1} + 2\binom{n}{2} + 3\binom{n}{3} + \dots + n\binom{n}{n} \]

We know \[ (1+x)^n = \binom{n}{0} + \binom{n}{1}x + \binom{n}{2}x^2 + \dots + \binom{n}{n}x^n \] Differentiate both sides \[ n(1+x)^{n-1} = \binom{n}{1} + 2\binom{n}{2}x + 3\binom{n}{3}x^2 + \dots + n\binom{n}{n}x^{n-1} \] Putting \(x=2\) in \[ n \cdot 2^{n-1} = \binom{n}{1} + 2\binom{n}{2} + 3\binom{n}{3} + \dots + n\binom{n}{n} \]

We know \[ (1+x)^n = \binom{n}{0} + \binom{n}{1}x + \binom{n}{2}x^2 + \dots + \binom{n}{n}x^n \] Differentiate both sides \[ n(1+x)^{n-1} = \binom{n}{1} + 2\binom{n}{2}x + 3\binom{n}{3}x^2 + \dots + n\binom{n}{n}x^{n-1} \] Putting \(x=-1\) in \[ n \cdot 2^{n-1} = \binom{n}{1} + 2\binom{n}{2} + 3\binom{n}{3} + \dots + n\binom{n}{n} \]

We know \[ (1+x)^n = \binom{n}{0} + \binom{n}{1}x + \binom{n}{2}x^2 + \dots + \binom{n}{n}x^n \] Multiply with \(x^3\) and differentiate w.r.t. \(x\): \[ x^3(1+x)^n = \binom{n}{0}x^3 + \binom{n}{1}x^4 + \binom{n}{2}x^5 + \dots \] Differentiate \[ 3x^2(1+x)^n + nx^3(1+x)^{n-1}= 3\binom{n}{0}x^2 + 4\binom{n}{1}x^3 + 5\binom{n}{2}x^4 + \dots \] Now put \(x=1\) \[ 3 \cdot 2^n + n \cdot 2^{n-1} = 3\binom{n}{0} + 4\binom{n}{1} + 5\binom{n}{2} + \dots + (n+3)\binom{n}{n} \]

\[ (1+x)^n = \binom{n}{0} + \binom{n}{1}x + \binom{n}{2}x^2 + \dots \]

Multiply by \(x\): \[ x(1+x)^n = \binom{n}{0}x + \binom{n}{1}x^2 + \binom{n}{2}x^3 + \dots \]

Differentiate w.r.t \(x\): \[ (1+x)^n + nx(1+x)^{n-1} = \binom{n}{0} + 2\binom{n}{1}x + 3\binom{n}{2}x^2 + \dots \]

Differentiate again w.r.t \(x\): \[ n(1+x)^{n-1} + n(1+x)^{n-1} + n(n-1)x(1+x)^{n-2} = 1 \cdot 2 \binom{n}{1} + 2 \cdot 3 \binom{n}{2}x + 3 \cdot 4 \binom{n}{3}x^2 + \dots \]

Put \(x=1\): \[ 2n \cdot 2^{n-1} + n(n-1)2^{n-2} = 1 \cdot 2 \binom{n}{1} + 2 \cdot 3 \binom{n}{2} + 3 \cdot 4 \binom{n}{3} + \dots \]

Hint: Multiply with \(x\) and differentiate. Do this twice.

\[ n(1+x)^{n-1} + n(1+x)^{n-1} + n(n-1)x(1+x)^{n-2} = 1 \cdot 2 \binom{n}{1} + 3 \cdot 4 \binom{n}{3}x^2 + 5 \cdot 6 \binom{n}{5}x^4 + \dots \]

Put \(x=1\) \[ 1 \cdot 2 \binom{n}{1} + 2 \cdot 3 \binom{n}{2} + 3 \cdot 4 \binom{n}{3} + \dots = 2n2^{n-1} + n(n-1)2^{n-2} \]

Put \(x=-1\) \[ 1 \cdot 2 \binom{n}{1} - 2 \cdot 3 \binom{n}{2} + 3 \cdot 4 \binom{n}{3} - \dots = 0 \]

Sum the above two equations \[ 2(1 \cdot 2 \binom{n}{1} + 3 \cdot 4 \binom{n}{3} + \dots) = 2n2^{n-1} + n(n-1)2^{n-2} \]

\[ \Rightarrow 1 \cdot 2 \binom{n}{1} + 3 \cdot 4 \binom{n}{3} + \dots = n2^{n-1} + n(n-1)2^{n-2} \]

\[ (1+x)^n = \binom{n}{0} + \binom{n}{1}x + \binom{n}{2}x^2 + \binom{n}{3}x^3 + \dots \]

As difference in successive coefficients is 3, \[ (1+x^3)^n = \binom{n}{0} + \binom{n}{1}x^3 + \binom{n}{2}x^6 + \binom{n}{3}x^9 + \dots \]

Multiply with \(x^7\) \[ x^7(1+x^3)^n = x^7 \binom{n}{0} + x^{10}\binom{n}{1} + \dots \]

Differentiate w.r.t \(x\) \[ 7x^6(1+x^3)^n + n(1+x^3)^{n-1} \cdot x^2 \cdot 7 = 7 \binom{n}{0}x^6 + \dots \]

Put \(x=1\) \[ 7 \cdot 2^n + n \cdot 2^{n-1} = 7\binom{n}{0} + 10\binom{n}{1} + 13\binom{n}{2} + \dots \]

\[ (1+x)^n = \binom{n}{0} + \binom{n}{1}x + \binom{n}{2}x^2 + \dots \]

\[ (1+x)^n = \binom{n}{0} + \binom{n}{1}x + \binom{n}{2}x^2 + \dots \]

Multiply the two equations \[ (1+x)^{2n} = \left(\binom{n}{0} + \binom{n}{1}x + \binom{n}{2}x^2 + \dots\right) \cdot \left(\binom{n}{0} + \binom{n}{1}x + \binom{n}{2}x^2 + \dots\right) \]

\[ (1+x)^{2n} = x^n \left( \binom{n}{0}\binom{n}{n} + \binom{n}{1}\binom{n}{n-1} + \dots \right) + \text{something} \]

We require coefficient corresponding to \(x^n\) \[ \Rightarrow\binom{n}{0}\binom{n}{n} + \binom{n}{1}\binom{n}{n-1} + \binom{n}{2}\binom{n}{n-2} + \dots = \binom{2n}{n} \]

Using same method as last question \[ (1+x)^{2n} = x^n \left( \binom{n}{0}\binom{n}{n} + \binom{n}{1}\binom{n}{n-1} + \dots \right) + x^{n-2} \left( \binom{n}{0}\binom{n}{n-2} + \binom{n}{1}\binom{n}{n-3} + \binom{n}{2}\binom{n}{n-4} + \dots + \binom{n}{n-2}\binom{n}{0} \right)+ \text{something} \] We need coefficient of \(x^{n-2}\) \[ \Rightarrow \binom{n}{0}\binom{n}{n-2} + \binom{n}{1}\binom{n}{n-3} + \binom{n}{2}\binom{n}{n-4} + \dots = \binom{2n}{n-2} \]

We start with binomial expansions: \[ (1-x)^n = \binom{n}{0} - \binom{n}{1}x + \binom{n}{2}x^2 - \binom{n}{3}x^3 + \cdots \] \[ (1+x)^n = \binom{n}{0} + \binom{n}{1}x + \binom{n}{2}x^2 + \binom{n}{3}x^3 + \cdots \]

Multiplying, \[ (1-x)^n(1+x)^n = (1-x^2)^n \] \[ = \Big(\binom{n}{0} - \binom{n}{1}x + \binom{n}{2}x^2 - \cdots\Big)\Big(\binom{n}{0} + \binom{n}{1}x + \binom{n}{2}x^2 + \cdots\Big) \]

We want the coefficient of \(x^{n-2}\). From the binomial theorem: \[ T_r = (-1)^r \binom{n}{r} (x^2)^r \]

Thus, \(x^{n-2}\) term appears when \[ 2r = n-2 \quad \Rightarrow \quad r = \frac{n-2}{2} \]

• If \(n\) is even: \(r\) is an integer, so the coefficient is \[ (-1)^{\tfrac{n-2}{2}} \binom{n}{\tfrac{n-2}{2}} \]
• If \(n\) is odd: \(\tfrac{n-2}{2}\) is not an integer, so no such term exists.

Recall: \[ (1+x)^n = \binom{n}{0} + \binom{n}{1}x + \binom{n}{2}x^2 + \cdots \tag{1} \] Differentiate w.r.t \(x\): \[ n(1+x)^{n-1} = \binom{n}{1} + 2\binom{n}{2}x + 3\binom{n}{3}x^2 + \cdots \tag{2} \]

Multiplying (1) and (2), \[ n(1+x)^{2n-1} = (\binom{n}{0} + \binom{n}{1}x + \binom{n}{2}x^2 + \cdots)(\binom{n}{1} + 2\binom{n}{2}x + 3\binom{n}{3}x^2 + \cdots) \]

Thus the coefficient of \(x^n\) is: \[ \binom{n}{1}\binom{n}{n} + 2\binom{n}{2}\binom{n}{n-1} + 3\binom{n}{3}\binom{n}{n-2} + \cdots + n\binom{n}{n}\binom{n}{1} \]

But from binomial theorem: \[ \text{Coefficient of } x^n \text{ in } n(1+x)^{2n-1} = n \binom{2n-1}{n} \]

Hence, \[ \binom{n}{1}\binom{n}{n} + 2\binom{n}{2}\binom{n}{n-1} + \cdots + n\binom{n}{n}\binom{n}{1} = n \binom{2n-1}{n} \]

Start with \[ (1+x)^n = \binom{n}{0} + \binom{n}{1}x + \binom{n}{2}x^2 + \cdots \tag{i} \] Multiply by \(x^3\) and differentiate: \[ \frac{d}{dx}\big[x^3(1+x)^n\big] = 3x^2(1+x)^n + nx^3(1+x)^{n-1} \] \[ = 3\binom{n}{0}x^2 + 4\binom{n}{1}x^3 + \cdots \tag{ii} \]

Multiply \((i)\) and \((ii)\): \[ 3x^2(1+x)^{2n} + nx^3(1+x)^{2n-1}= (3\binom{n}{0}x^2 + 4\binom{n}{1}x^3 + \cdots)(\binom{n}{0} + \binom{n}{1}x + \binom{n}{2}x^2 + \cdots) \]

So, we need the coefficient of \(x^{n+1}\): \[ 3\binom{2n}{n-1} + n\binom{2n-1}{n-2} \]

Let \[ S = \binom{n}{0}\binom{n}{n} - \binom{n}{1}\binom{n+1}{n} + \binom{n}{2}\binom{n+2}{n} - \cdots \]

We note that \[ \binom{n}{r}\binom{n+r}{n} = \text{coefficient of } x^n \text{ in } \binom{n}{r}(1+x)^{n+r} \]

Hence \[ S = \text{coefficient of } x^n \text{ in } \bigg[ \binom{n}{0}(1+x)^n - \binom{n}{1}(1+x)^{n+1} + \binom{n}{2}(1+x)^{n+2} - \cdots \bigg] \]

Factorizing: \[ S = \text{coefficient of } x^n \text{ in } (1+x)^n \bigg[\binom{n}{0} - \binom{n}{1}(1+x) + \binom{n}{2}(1+x)^2 - \cdots\bigg] \]

The expression inside brackets is: \[ (1 - (1+x))^n = (-x)^n \]

\[ \therefore S = \text{coefficient of } x^n \text{ in } (1+x)^n \cdot (-x)^n \]

Simplify: \[ (1+x)^n \cdot (-x)^n = (-1)^n x^n (1+x)^n \]

The coefficient of \(x^n\) is: \[ (-1)^n \binom{n}{n} = (-1)^n. \] \[ \Rightarrow S = (-1)^n \]

Method 1:

Start with: \[ (1+x^2)^{2n} = \binom{2n}{0} + \binom{2n}{1}x^2 + \binom{2n}{2}x^4 + \cdots + \binom{2n}{2n}x^{4n} \]

Multiply by \(x\): \[ x(1+x)^{2n} = \binom{2n}{0}x + \binom{2n}{1}x^3 + \binom{2n}{2}x^5 + \cdots + \binom{2n}{2n}x^{4n+1} \]

Differentiate w.r.t \(x\): \[ \frac{d}{dx}\Big[x(1+x^2)^{2n}\Big] = (1+x^2)^{2n} + 2n x (1+x^2)^{2n-1}=\binom{2n}{0} + 3\binom{2n}{1}x^2 + \cdots \]

Putting \(x=1\): \[ 2^{2n} + n \cdot 2^{2n-1} = \binom{2n}{0} + 3\binom{2n}{1} + \cdots + (2n+1) \text{ terms} \]

Method 2:

Consider the general term: \[ T_r = (2r+1)^2\binom{2n}{r}=4r^2\binom{2n}{r}+\binom{2n}{r}+4r\binom{2n}{r} \] Now calculate the sum of each term individually.

coeff of \(x^n\) in \[ \big[(1+x)^n + (1+x)^{n+1} + \cdots + (1+x)^{2n}\big] \]

\[= (1+x)^n \Big[1 + (1+x) + \cdots + (1+x)^n\Big]\] Solve the term inside bracket using sum of geometric series. Then, identify the coefficient of \(x^n\).

\[f(n)= \sum \Big((r+1)^2 \binom{n}{r} - (r)^2 \binom{n}{r-1}\Big)\]

Use \(V_n\) method from sequence and series to solve this.