MCQ
$\sum\limits_{k = 0}^{10} {^{20}{C_k} = } $
  • ${2^{19}} + \frac{1}{2}{\,^{20}}{C_{10}}$
  • B
    ${2^{19}}$
  • C
    $^{20}{C_{10}}$
  • D
    None of these

Answer

Correct option: A.
${2^{19}} + \frac{1}{2}{\,^{20}}{C_{10}}$
a
(a) $\sum\limits_{K = 0}^{10} {^{20}{C_k}} $ i.e., $^{20}{C_0} + {\,^{20}}{C_1} + ...... + {\,^{20}}{C_{10}}$

We know that, ${(1 + x)^n} = {\,^n}{C_0} + {\,^n}{C_1}{x^1} + {\,^n}{C_2}{x^2} + .... + {\,^n}{C_n}.{x^n}$

Put $x = 1$; ${2^n} = {\,^n}{C_0} + {\,^n}{C_1} + {\,^n}{C_2} + ..... + {\,^n}{C_n}$

Put $n = 20$; ${2^{20}} = {\,^{20}}{C_0} + {\,^{20}}{C_1} + {\,^{20}}{C_2} + ...... + {\,^{20}}{C_{20}}$

${2^{20}} + \,{\,^{20}}{C_{10}} = 2\,[{\,^{20}}{C_0} + {\,^{20}}{C_1} + ...... + {\,^{20}}{C_{10}}]$

${[^{20}}{C_0} + {\,^{20}}{C_1} + ...... + {\,^{20}}{C_{10}}] = {2^{19}} + \frac{1}{2}{\,^{20}}{C_{10}}$

$\sum\limits_{k = 0}^{10} {^{20}{C_k}} = {2^{19}} + \frac{1}{2}{\,^{20}}{C_{10}}$.

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