Binomial Theorem

This lecture notes on binomial theorem is designed for advanced high school and intermediate students preparing for competitive examinations.

defines binomial coefficients and expansion. contains solved problems on series involving binomial coefficients. describes multinomial expansion, a generalization of binomial expansion.

Binomial expansion and coefficients

Binomial theorem

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

Binomial coefficients

\[ \binom{n}{r} = \frac{n!}{(n-r)! \, r!} = \binom{n}{n-r} \] \[ = \frac{n (n-1) (n-2) \dots (n-r+1)}{r!} \]

Symmetric binomial coefficients

If $\binom{n}{r_1} = \binom{n}{r_2}$, then either $r_1 = r_2$ or $r_1 = n-r_2$.

The $n^{th}$ term in the binomial expansion is \[ T_{r+1} = \binom{n}{r} x^{n-r} a^r \] The $x$-th term from last is $T_{(n-x)+2}$.

Middle term in binomial expansion

If $n$ is even, middle term = $T_{\frac{n}{2}+1}$
If $n$ is odd, middle terms = $T_{\frac{n+1}{2}}$ and $T_{\frac{n+3}{2}}$

Binomial coefficient is maximum for middle term(s)

Find $7^{th}$ term in the binomial expansion of $(3+x)^9$.

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

Find coeff. of $x^2$ in $(x-2)^2 (x+7)^{13}$.

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

Find coeff. of $x^2$ in $(x-2)^2 (x-7)^{13}$.

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

Find coeff. of $x^{15}$ in $(x^2 - \tfrac{1}{x})^{51}$.

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} $

Find coeff. of $x^{11}$ in $(x^2 + 3x + 2)^6$.

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

Find coeff. of $x^r$ where $r \in [0,n]$ in $(1+x) + (1+x)^2 + (1+x)^3 + \dots + (1+x)^n$.

\[ S = (1+x) + (1+x)^2 + \dots + (1+x)^n \] 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. \[ \Rightarrow \binom{n+1}{r+1} x^r \]

Numerically greatest term (maximum modulus term) in binomial expansion

Let greatest term be $T_{r+1}$, such that \[ |T_{r+1}| \geq |T_r| \] \[ \left|\binom{n}{r} x^{n-r} a^r\right| \geq \left|\binom{n}{r-1} x^{n-r+1} a^{r-1}\right| \] \[ \Rightarrow \frac{\binom{n}{r}}{\binom{n}{r-1}} |a| \geq |x| \] \[ \Rightarrow \frac{n-r+1}{r} |a| \geq |x| \] \[ \frac{|a|}{r} \geq \frac{|x|}{n-r+1} \] \[ \Rightarrow \frac{(n+1)|a|}{|x|+|a|} > r \] Hence, till the time this inequality is satisfied, value of $r$ can be found for the greatest numerical term.

Case I: If inequality does not hold in integer, then \[ r = \lfloor \text{LHS} \rfloor + 1 \] Example: For 4.53, answer = 5.

Case II: If inequality holds in integer, then \[ r = \text{LHS}, \quad r = \text{LHS} + 1 \] Example: For 4, answers are 4 and 5. Thus, both $T_5$ and $T_4$ are numerically greatest.

Properties of binomial coefficients

Properties of binomial coefficients.

Put $x=1$ in the binomial expansion of $(1+x)^n$ \[ \binom{n}{0} + \binom{n}{1} + \binom{n}{2} + \dots + \binom{n}{n}=2^n \] Put $x=-1$ in the binomial expansion of $(1+x)^n$ \[ \binom{n}{0} - \binom{n}{1} + \binom{n}{2} - \binom{n}{3} + \dots=0 \] Together they imply \[\binom{n}{0} + \binom{n}{2} + \binom{n}{4} + \dots \dots = 2^{n-1} =\binom{n}{1} + \binom{n}{3} + \binom{n}{5}+ \dots + \dots \]

Binomial coefficients identities

\[\binom{n}{r} = \frac{n}{r} \binom{n-1}{r-1}\] \[\binom{n}{r} + \binom{n}{r+1} = \binom{n+1}{r+1}\]

Proof

\[ \binom{n}{r} = \frac{n!}{(n-r)!r!} = \frac{n}{r} \cdot \frac{(n-1)!}{(n-r)!(r-1)!} = \frac{n}{r} \binom{n-1}{r-1} \]

\[ \binom{n}{r} + \binom{n}{r+1} = \frac{n!}{(n-r)!r!} + \frac{n!}{(n-r-1)!(r+1)!} = \frac{n!(r+1) + n!(n-r)}{(n-r)!(r+1)!} = \frac{(n+1)!}{(n-r)!(r+1)!} = \binom{n+1}{r+1} \]

Series containing binomial coefficients

Calculate the sum of the series\[\binom{n}{1} + 2\binom{n}{2} + 3\binom{n}{3} + \dots + n\binom{n}{n}\]

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 $x=1$ \[ n \cdot 2^{n-1} = \binom{n}{1} + 2\binom{n}{2} + 3\binom{n}{3} + \dots + n\binom{n}{n} \]

Calculate the sum of the series\[\binom{n}{1} - 2\binom{n}{2}2 + 3\binom{n}{3}2^2 - \dots + n\binom{n}{n}2^n\]

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 \[ n \cdot 2^{n-1} = \binom{n}{1} + 2\binom{n}{2} + 3\binom{n}{3} + \dots + n\binom{n}{n} \]

Calculate the sum of the series\[\binom{n}{1} - 2\binom{n}{2} + 3\binom{n}{3} - \dots \]

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 \[ n \cdot 2^{n-1} = \binom{n}{1} + 2\binom{n}{2} + 3\binom{n}{3} + \dots + n\binom{n}{n} \]

Calculate the sum of the series\[3\binom{n}{0} + 4\binom{n}{1} + 5\binom{n}{2} + 6\binom{n}{3} + \dots + (n+3)\binom{n}{n}\]

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 $x=1$ \[ 3 \cdot 2^n + n \cdot 2^{n-1} = 3\binom{n}{0} + 4\binom{n}{1} + 5\binom{n}{2} + \dots + (n+3)\binom{n}{n} \]

Calculate the sum of the series\[ 1 \cdot 2 \binom{n}{1} + 2 \cdot 3 \binom{n}{2} + \dots \]

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

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

Calculate the sum of the series\[1^2 \cdot 2^2 \binom{n}{1} + 2^2 \cdot 3^2 \binom{n}{2} + \dots \]

Hint: Multiply with $x$ and differentiate. Do this twice.

Calculate the sum of the series\[1 \cdot 2 \binom{n}{1} + 3 \cdot 4 \binom{n}{3} + 5 \cdot 6 \binom{n}{5} + \dots \]

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

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

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

Calculate the sum of the series \[7 \binom{n}{0} + 10 \binom{n}{1} + 13 \binom{n}{2} + 16 \binom{n}{3} + \dots \]

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

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 $x$ \[ 7x^6(1+x^3)^n + n(1+x^3)^{n-1} \cdot 3 x^2 \cdot x^7 = 7 \binom{n}{0}x^6 + \dots \]

Put $x=1$ \[ 7 \cdot 2^n + 3n \cdot 2^{n-1} = 7\binom{n}{0} + 10\binom{n}{1} + 13\binom{n}{2} + \dots \]

Series with product of two binomial coefficients having constant upper suffix and constant sum of lower suffix

Calculate the sum of the series \[\binom{n}{0}\binom{n}{n} + \binom{n}{1}\binom{n}{n-1} + \binom{n}{2}\binom{n}{n-2} + \dots + \binom{n}{n}\binom{n}{0} \]

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

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 $x^n$ \[ \Rightarrow\binom{n}{0}\binom{n}{n} + \binom{n}{1}\binom{n}{n-1} + \binom{n}{2}\binom{n}{n-2} + \dots = \binom{2n}{n} \]

Calculate the sum of the series \[\binom{n}{0}\binom{n}{n-2} + \binom{n}{1}\binom{n}{n-3} + \binom{n}{2}\binom{n}{n-4} + \dots + \binom{n}{n-2}\binom{n}{0}\]

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 $x^{n-2}$ \[ \Rightarrow \binom{n}{0}\binom{n}{n-2} + \binom{n}{1}\binom{n}{n-3} + \binom{n}{2}\binom{n}{n-4} + \dots = \binom{2n}{n-2} \]

Calculate the sum of the series \[ \binom{n}{0}\binom{n}{n-2} - \binom{n}{1}\binom{n}{n-3} + \binom{n}{2}\binom{n}{n-4} - \dots + (-1)^{n-2}\binom{n}{n-2}\binom{n}{0} \]

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.

Calculate the sum of the series \[ \binom{n}{1}\binom{n}{n} + 2\binom{n}{2}\binom{n}{n-1} + 3\binom{n}{3}\binom{n}{n-2} + \cdots + n\binom{n}{n}\binom{n}{1} \]

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

Calculate the sum of the series \[ 3\binom{n}{0}\binom{n}{n-1} + 4\binom{n}{1}\binom{n}{n-2} + \cdots \]

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 $x^{n+1}$: \[ 3\binom{2n}{n-1} + n\binom{2n-1}{n-2} \]

Find the sum of \[ \binom{n}{0}\binom{n}{n} - \binom{n}{1}\binom{n+1}{n} + \binom{n}{2}\binom{n+2}{n} - \cdots \text{ up to $n$ terms} \]

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: \[ (-1)^n \binom{n}{n} = (-1)^n. \] \[ \Rightarrow S = (-1)^n \]

Evaluate \[ \sum_{r=0}^n (2r+1)\binom{2n}{r} \]

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.

Calculate \[ \binom{n}{n} + \binom{n+1}{n} + \cdots + \binom{2n}{n} \]

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 $x^n$.

$f(n) = \sum_{r=1}^n \left[ (r)^2 \left(\binom{n}{r}-\binom{n}{r-1}\right) + (2r+1)\binom{n}{r}\right]$. Find $\frac{f(30)}{120}$

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

Use $V_n$ method from sequence and series to solve this.

Multinomial theorem

Coefficient of $x^\alpha y^\beta z^\gamma$ in $(x+y+z)^n$, where $\alpha + \beta + \gamma = n$, is \[ \frac{n!}{\alpha! \, \beta! \, \gamma!} \]

\[(x+y+z)(x+y+z)\cdots(x+y+z)\quad n \text{ times}\]

$\alpha$–$x$ has to be taken out of $n$ brackets $\Rightarrow \binom{n}{\alpha}$
$\beta$–$y$ has to be taken out of remaining brackets $= \binom{n-\alpha}{\beta}$
$\gamma$–$z$ has to be taken out of remanining brackets $= \binom{n-\alpha-\beta}{\gamma}$

Therefore, the coefficient is \[\binom{n}{\alpha} \binom{n-\alpha}{\beta} \binom{n-\alpha-\beta}{\gamma}\]

\[=\dfrac{n!}{\alpha! (n-\alpha)!} \cdot \dfrac{(n-\alpha)!}{\beta! (n-\alpha-\beta)!} \cdot \dfrac{(n-\alpha-\beta)!}{\gamma!}\]

\[=\dfrac{n!}{\alpha! \, \beta! \, \gamma!}\]

Author

Anurag Gupta is an M.S. graduate in Electrical and Computer Engineering from Cornell University. He also holds an M.Tech degree in Systems and Control Engineering and a B.Tech degree in Electrical Engineering from the Indian Institute of Technology, Bombay.

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