The sum of the last eight consecutive coeffic | vedclass

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(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

Let $X = 1({ }^{10} C _1)^2 + 2({ }^{10} C _2)^2 + 3({ }^{10} C _3)^2 + \ldots + 10({ }^{10} C _{10})^2$,where ${ }^{10} C _{ r }$ for $r \in \{1, 2, \ldots, 10\}$ denotes binomial coefficients. Then,the value of $\frac{1}{1430} X$ is:

$^{15}C_3 + ^{15}C_5 + \ldots + ^{15}C_{15} = ?$

If $n$ is a positive integer,then $\sum_{r=1}^n r^2 \cdot C_r = (\ldots \ldots \ldots) 2^{n-2}$

Statement $-1$: $\sum_{r=0}^{n} (r+1) \binom{n}{r} = (n+2) 2^{n-1}$
Statement $-2$: $\sum_{r=0}^{n} (r+1) \binom{n}{r} x^r = (1+x)^n + nx(1+x)^{n-1}$

$\sum\limits_{k = 0}^{10} {^{20}{C_k} = }$

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