If the expression 5^{2n} - 48n + k is divisib | vedclass

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(D) Let $f(n) = 5^{2n} - 48n + k = 25^n - 48n + k$.We can write $25^n$ as $(1 + 24)^n$.Using the Binomial Theorem,$(1 + 24)^n = 1 + n(24) + \frac{n(n-

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$23$

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(D) Let $f(n) = 5^{2n} - 48n + k = 25^n - 48n + k$.We can write $25^n$ as $(1 + 24)^n$.Using the Binomial Theorem,$(1 + 24)^n = 1 + n(24) + \frac{n(n-1)}{2}(24)^2 + \dots + 24^n$.So,$25^n = 1 + 24n + 24^2 	imes \frac{n(n-1)}{2} + \dots + 24^n$.Substituting this into the expression,we get $f(n) = 1 + 24n + 24^2 	imes \frac{n(n-1)}{2} + \dots + 24^n - 48n + k$.$f(n) = 1 + 24n - 48n + k + 24^2 	imes \frac{n(n-1)}{2} + \dots + 24^n$.$f(n) = 1 - 24n + k + 24^2 	imes \frac{n(n-1)}{2} + \dots + 24^n$.For $f(n)$ to be divisible by $24$ for all $n \in N$,the term $(1 - 24n + k)$ must be divisible by $24$.Since $24n$ is already divisible by $24$,$(1 + k)$ must be divisible by $24$.Thus,$1 + k = 24m$ for some integer $m$.For the least positive integral value of $k$,we set $m = 1$,which gives $1 + k = 24$,so $k = 23$.

If the expression $5^{2n} - 48n + k$ is divisible by $24$ for all $n \in N$,then the least positive integral value of $k$ is

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