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STD 11 - 7. binomial theoram — JEE STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceJEESTD 11 - 7. binomial theoram4 Marks
MCQ
The remainder left out when ${8^{2n}} - {\left( {62} \right)^{2n + 1}}$ is divided by $9$ is :
  • ✓
    $2$
  • B
    $7$
  • C
    $8$
  • D
    $0$
✓

Answer

Correct option: A.
$2$
a
$8^{2 n}-(62)^{2 n+1}$

$=(1+63)^{n}-(63-1)^{2 n+1}$ 

$=(1+63)^{n}+(1-63)^{2 n+1}$

$=\left(1+^{n} C_{1} 63+^{n} C_{2}(63)^{2}+\ldots+(63)^{n}\right.$

$+\left[1-^{(2 n+1)} C_{1} 63+(2 n+1) C_{2}(63)^{2}\right.$

$\left.+\ldots+(-1)(63)^{(2 n+1)}\right] $

$=2+63\left[^{n} C_{1}+^{n} C_{2}(63)+\ldots+(63)^{n-1}\right] $

$\left.+^{(2 n+1)} C_{2}(63)-\ldots-(63)^{(2 n)}\right]$

Hence, the remainder is $2.$

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