: Sum of the series S=3−1 10C0−32 10C1+32 10C2−33 10C3+....…+310 10C10 is

A
$2^{9}$
B
$2^{10} - 1$
C
$\frac{1}{3} (2^{11} - 2)$
D
$\frac{1}{3} (2^{10})$

Solution:

We have

$\begin{array}{l} S = 3^{- 1} [^{10} C_{0} -^{10} C_{1} (3) +^{10} C_{2} (3^{2}) -^{10} C_{3} (3^{3}) + \dots . . . . . +^{10} C_{10} (3^{10})] \\ = \frac{1}{3} (1 - 3)^{10} = \frac{1}{3} (2^{10}) \end{array}$

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