First slide

Binomial theorem for positive integral Index

Question

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

Moderate

A

$2^{16}$
B

$2^{15}$
C

$2^{14}$
D

none of these

Solution

We have,
$(1 + x)^{15} =^{15} C_{0} +^{15} C_{1} x +^{15} C_{2} x^{2} + \dots +^{15} C_{15} x^{15}$
We have to find the value of
$\begin{array}{l} ^{15} C_{8} +^{15} C_{9} +^{15} C_{10} + \dots +^{15} C_{15} \\ = \frac{1}{2} \{2 \times^{15} C_{8} + 2 \times^{15} C_{9} + 2 \times^{15} C_{10} + \dots + 2 \times^{15} C_{15}\} \end{array}$
$= \frac{1}{2} \{(^{15} C_{7} +^{15} C_{8}) + (^{15} C_{6} +^{15} C_{9}) + (^{15} C_{5} +^{15} C_{10}) + \dots$
$\begin{array}{l} + \dots \dots \dots \dots . . + (^{15} C_{0} +^{15} C_{15})\} [∵^{n} C_{r} =^{n} C_{n - r}] \\ = \frac{1}{2} (^{15} C_{0} +^{15} C_{1} +^{15} C_{2} + \dots +^{15} C_{15}) = \frac{1}{2} \times 2^{15} = 2^{14} \end{array}$

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