The expression {49^n} + 16n - 1 is divisible | vedclass

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Question

(C) We use the binomial expansion: ${49^n} = {(1 + 48)^n}$.Using the binomial theorem,${49^n} = 1 + n(48) + \frac{n(n-1)}{2}(48^2) + \dots + 48^n$.Sub

Vedclass · Accepted Answer

$64$

Vedclass · Answer

(C) We use the binomial expansion: ${49^n} = {(1 + 48)^n}$.Using the binomial theorem,${49^n} = 1 + n(48) + \frac{n(n-1)}{2}(48^2) + \dots + 48^n$.Substituting this into the expression:${49^n} + 16n - 1 = (1 + 48n + \frac{n(n-1)}{2}(2304) + \dots) + 16n - 1$.$= 48n + 16n + \frac{n(n-1)}{2}(2304) + \dots$$= 64n + 1152n(n-1) + \dots$Since $1152 = 64 	imes 18$,every term in the expansion is a multiple of $64$.Thus,${49^n} + 16n - 1$ is divisible by $64$.

The expression ${49^n} + 16n - 1$ is divisible by:

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