Using binomial theorem, prove that 23n – 7n – 1 is divisible by 49, where n ∈ N.

Using binomial theorem, prove that 23n – 7n – 1 is divisible by 49, where n ∈ N.

Asked by Aaryan | 1 year ago |  75
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Class 11 Maths Binomial Theorem         Add Answer

1 Answer

Solution :-

Given:

23n – 7n – 1

So, 23n – 7n – 1 = 8n – 7n – 1

Now,

8n – 7n – 1

8n = 7n + 1

= (1 + 7) n

= nC0 + nC1 (7)1 + nC2 (7)2 + nC3 (7)3 + nC4 (7)2 + nC5 (7)1 + … + nCn (7) n

8n = 1 + 7n + 49 [nC2 + nC3 (71) + nC4 (72) + … + nCn (7) n-2]

8n – 1 – 7n = 49 (integer)

So now,

8n – 1 – 7n is divisible by 49

Or

23n – 1 – 7n is divisible by 49.

Hence proved.

Answered by Aaryan | 1 year ago
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