The sum of the last eight consecutive coeffic | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Let $S$ be the sum of the last eight coefficients in the expansion of $(1+x)^{15}$.$S = \binom{15}{8} + \binom{15}{9} + \binom{15}{10} + \binom{15

Vedclass · Accepted Answer

$2^{14}$

Vedclass · Answer

(B) Let $S$ be the sum of the last eight coefficients in the expansion of $(1+x)^{15}$.$S = \binom{15}{8} + \binom{15}{9} + \binom{15}{10} + \binom{15}{11} + \binom{15}{12} + \binom{15}{13} + \binom{15}{14} + \binom{15}{15}$  .......$(i)$Using the property $\binom{n}{r} = \binom{n}{n-r}$,we have $\binom{15}{8} = \binom{15}{7}$,$\binom{15}{9} = \binom{15}{6}$,...,$\binom{15}{15} = \binom{15}{0}$.Thus,$S = \binom{15}{7} + \binom{15}{6} + \binom{15}{5} + \binom{15}{4} + \binom{15}{3} + \binom{15}{2} + \binom{15}{1} + \binom{15}{0}$  .......$(ii)$Adding $(i)$ and $(ii)$:$2S = \sum_{k=0}^{15} \binom{15}{k} = 2^{15}$$S = \frac{2^{15}}{2} = 2^{14}$

The sum of the last eight consecutive coefficients in the expansion of $(1+x)^{15}$ is

Similar Questions

If $(\frac{1}{^{15}C_{0}}+\frac{1}{^{15}C_{1}})(\frac{1}{^{15}C_{1}}+\frac{1}{^{15}C_{2}})...(\frac{1}{^{15}C_{12}}+\frac{1}{^{15}C_{13}}) = \frac{a^{13}}{^{14}C_{0} \cdot ^{14}C_{1} \cdot ... \cdot ^{14}C_{12}}$, then $30a$ is equal to:

If $\sum_{k=1}^{10} k^{2} \binom{10}{k}^{2} = 22000 L$,then $L$ is equal to $.....$

If $(1+x+x^2)^n = a_0 + a_1 x + a_2 x^2 + \ldots + a_{2n} x^{2n}$,then $a_0 + a_2 + a_4 + \ldots + a_{2n} =$

Suppose $\sum_{r=0}^{2023} r \cdot ^{2023}C_r = 2023 \times \alpha \times 2^{2022}$. Then the value of $\alpha$ is $............$

If $C_r = { }^n C_r$,then find the sum $C_0 + C_4 + C_8 + C_{12} + \ldots$

Vedclass Test Series

Exam Paper Generator

Online Exam Module