What is the sum of the coefficients?

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Hint: Sum of coefficients of ${\left( {x + y} \right)^n}$ is obtained when we put $x = y = 1$. And the greatest coefficient is the coefficient of the middle term(s) in its binomial expansion.According to the question, the sum of coefficients in the expansion of ${\left( {x + y} \right)^n}$ is 4096. We know that the sum of coefficients is the value of the expansion if we put all the variables equal to 1. Hence here we will put $x = y = 1$. So, we have:$ \Rightarrow {\left( {1 + 1} \right)^n} = 4096, \\ \Rightarrow {2^n} = 4096, \\ \Rightarrow {2^n} = {2^{12}}, \\ \Rightarrow n = 12 \\ $Since $n = 12$, the expansion is of ${\left( {x + y} \right)^{12}}$ and it will have a total of 13 terms.We know that the greatest coefficient is the middle term. In this case, it will be of 7th term.The general term for binomial expansion of ${\left( {x + y} \right)^{12}}$ is:$ \Rightarrow {T_{r + 1}}{ = ^{12}}{C_r}{x^{12 - r}}.{y^r}$For middle term (i.e. 7th term), we will put $r = 6$:$ \Rightarrow {T_7}{ = ^{12}}{C_6}{x^6}.{y^6}$Thus the coefficient of the middle term is $^{12}{C_6} = 924$And hence the greatest coefficient in the expansion is 924.Note:In the expansion of ${\left( {x + y} \right)^n}$, coefficient of the middle term is $^n{C_{\dfrac{n}{2}}}$ if $n$ is even. But if $n$ is odd, there will be two middle terms having coefficients $^n{C_{\dfrac{{\left( {n - 1} \right)}}{2}}}$ and $^n{C_{\dfrac{{\left( {n + 1} \right)}}{2}}}$. The value of the coefficients will be the same though.

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(Last Updated On: January 21, 2020)

Problem Statement: ECE Board November 1995

What is the sum of the coefficients of the expansion (2x–1)^20?

Problem Answer:

The sum of the coefficients of the expansion is 0.

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Hint: Try a few more small cases and see what happens (keep using Pascal's triangle... note that the signs of each term alternate beginning with a positive sign).

$1, -2, 1 \rightarrow 0$ This is $n=2$.

$1, -3, 3, -1 \rightarrow 0$ This is $n=3$.

$1, -4, 6, -4, 1 \rightarrow 0$ This is $n=4$.

$1, -5, 10, -10, 5, -1 \rightarrow 0$ This is $n=5$.

$1, -6, 15, -20, 15, -6, 1 \rightarrow 0$ This is $n=6$.

The case when $n$ is odd is fairly easy because $\binom{n}{i}=\binom{n}{n-i}$ and exactly one of those will be positive and exactly one will be negative (so they cancel). What happens when $n$ is even is a little more subtle.

You would be done if you could show why $\sum \binom{n}{2i} = \sum \binom{n}{2i+1}$ when $n$ is even. There is a nice combinatorial way to do this, but I'll let you look for it.