MCQ
If $^{2017}C_0 + ^{2017}C_1 + ^{2017}C_2+......+ ^{2017}C_{1008} = \lambda ^2 (\lambda   > 0),$ then remainder when $\lambda $ is divided by $33$ is-
  • A
    $8$
  • B
    $13$
  • C
    $17$
  • $25$

Answer

Correct option: D.
$25$
d
${\,^{2017}}{C_0} + {\,^{2017}}{C_1} + ......{\,^{2017}}{C_{1008}} = {2^{2016}} = {\lambda ^2}$

$\lambda  = {2^{1008}}\,\,\,\,\, \Rightarrow \,\,\,{8.32^{201}} = 8{\left( {33 - 1} \right)^{201}} =  - 8 = 25$

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