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The remainder left out when 8^{2n}-\left ( 62 \right )^{2n+1} is divided by 9 is

Option: 1

2


Option: 2

7


Option: 3

8


Option: 4

0


Answers (1)

As we have elarned

Expression of Binomial Theorem -

\left ( x+a \right )^{n}= ^{n}\! C_{0}x^{n}a^{0}+^{n}C_{1}x^{n-1}a^{1}+^{n}C_{2}x^{n-2}a^{2}+-----+ ^{n}C_{n}x^{0}a^{n}

 

Now,

(8)^{2n}-(62)^{2n+1}\\ = (9-1)^{2n}- (63-1)^{2n+1}\\

= [ ^{2n}C_0 (9)^{2n}- ^{2n}C_1(9)^{2n-1}+....-^{2n}C_{2n-1}9+ ^{2n}C_{2n}]- [ ^{2n+1}C_0 (63)^{2n+1} \\- ^{2n+1}C_1(63)^{2n}+....+ ^{2n+1}C_{2n}63 - ^{2n+1 }C_{2n+1}]

= 9k  + ^{2n}C_{2n}+ ^{2n+1}C_{2n+1}

On dividing by 9 

Remainder = ^{2n}C_{2n}+ ^{2n+1}C_{2n+1} = 1 + 1 = 2

 

 

 

 

 

 

 

 

 

Posted by

Kshitij

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