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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.
Binomial Theorem | CBSE | Class 11 | Exercise 18.1 | RD Sharma
Using binomial theorem, prove that 23n – 7n – 1 is divisible by 49, where n ∈ N.

Answer:

Given,

23n – 7n – 1

23n – 7n – 1 = 8n – 7n – 1

Using binomial theorem,

8n = 7n + 1

8n = (1 + 7) n

8n = 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

8n – 1 – 7n is divisible by 49

23n – 1 – 7n is divisible by 49.

Thus, proved.

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