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

Asked by Aaryan | 1 year ago |  75

##### 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) + nC(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

### Related Questions

#### Find the term independent of x in the expansion of (3/2 x2 – 1/3x)9

Find the term independent of x in the expansion of $$(\dfrac{3}{2x^2} – \dfrac{1}{3x})^9$$

#### Find the middle term in the expansion of (x – 1/x)2n+1

Find the middle term in the expansion of $$(x-\dfrac{ 1}{x})^{2n+1}$$

#### Find the middle term in the expansion of (1 + 3x + 3x2 + x3)2n

Find the middle term in the expansion of (1 + 3x + 3x2 + x3)2n

Find the middle term in the expansion of $$(\dfrac{x}{a} – \dfrac{a}{x})^{10}$$