The least positive integer greater than 1 tha | vedclass

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Question

(A) Let $f(n) = 49^n + 16n - 1$. For $n = 1$,$f(1) = 49^1 + 16(1) - 1 = 49 + 16 - 1 = 64$. For $n = 2$,$f(2) = 49^2 + 16(2) - 1 = 2401 + 32 - 1 = 2432

Vedclass · Accepted Answer

$64$

Vedclass · Answer

(A) Let $f(n) = 49^n + 16n - 1$. For $n = 1$,$f(1) = 49^1 + 16(1) - 1 = 49 + 16 - 1 = 64$. For $n = 2$,$f(2) = 49^2 + 16(2) - 1 = 2401 + 32 - 1 = 2432$. We check if $64$ divides $f(2)$: $2432 / 64 = 38$. Since $64$ divides both $f(1)$ and $f(2)$,we test the expression using the binomial expansion: $49^n = (1 + 48)^n = 1 + n(48) + \frac{n(n-1)}{2}(48^2) + \dots$ $49^n = 1 + 48n + 1152n(n-1) + \dots$ Substituting this into $f(n)$: $f(n) = (1 + 48n + 1152n(n-1) + \dots) + 16n - 1$ $f(n) = 64n + 1152n(n-1) + \dots$ Since $1152 = 64 	imes 18$,every term in the expansion is divisible by $64$. Thus,the least positive integer greater than $1$ that divides $f(n)$ for all $n$ is $64$.

The least positive integer greater than $1$ that divides $49^n + 16n - 1$ for all positive integers $n$ is

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