Question

Sum of last $30$ coefficients in the binomial expansion of $(1 + x)^{59}$ is

• A. $2^{29}$
• B. $2^{59}$
• C. $2^{58}$
• D. $2^{59} - 2^{29}$

Answer 1

The Correct Option is C

We have, $(1+x)^{59}$ Sum of last 30 coefficient of the binomial expansion $={ }^{59} C_{30}+{ }^{59} C_{31}+\ldots+{ }^{59} C_{59}$ We know that, ${ }^{59} C_{0}+{ }^{59} C_{1}+{ }^{59} C_{2}+\ldots+{ }^{59} C_{59}= 2^{59}$ $\Rightarrow\left({ }^{59} C_{0}+{ }^{59} C_{59}\right)+\left({ }^{59} C_{1}+{ }^{59} C_{58}\right)+...$ $+\left({ }^{59} C_{29}+{ }^{59} C_{30}\right)=2^{59}$ $\Rightarrow 2\left({ }^{59} C_{59}+{ }^{59} C_{58}+...+{ }^{59} C_{31}+{ }^{59} C_{30}\right)=2^{59}$ $\left[\because{ }^{n} C_{r}={ }^{n} C_{n-r}\right]$ $\Rightarrow{ }^{59} C_{30}+{ }^{59} C_{31}+...{ }^{59} C_{39}=\frac{2^{59}}{2}=2^{58}$ $\therefore$ Sum of last 30 coefficient of the binomial expansion $(1+x)^{59}$ is $2^{58}$.

Answer 2

To find the sum of the last 30 coefficients in the binomial expansion of \((1 + x)^{59}\), we can use the formula for the binomial expansion:

\((1 + x)^{59} = \binom{59}{0} + \binom{59}{1}x + \binom{59}{2}x^2 + \ldots + \binom{59}{59}x^{59}\)

In this expansion, the coefficient of \(x^k\) term is given by \(\binom{59}{k}\).
We want to find the sum of the coefficients from \(x^{59}\) down to \(x^{30}\), inclusive.

So, the sum we need to find is:
\(\binom{59}{59} + \binom{59}{58} + \binom{59}{57} + \ldots + \binom{59}{30}\)

Now, recall that the sum of the coefficients in the expansion of \((1+x)^n\ is\ 2^n\).
So, in our case, the sum of all coefficients in the expansion of \((1+x)^{59}\ is \ 2^{59}\).

Therefore, the sum of the last 30 coefficients is half of the sum of all coefficients:

\(\frac{1}{2} \times 2^{59} = 2^{58}\)

So, the correct option is (C): \(2^{58}\)