If k in N,then 3^{3k} - 26k - 1 is divisible | vedclass

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(A) Let $f(k) = 3^{3k} - 26k - 1 = 27^k - 26k - 1$. Using the Binomial Theorem,we can write $27^k$ as $(1 + 26)^k$. $27^k = (1 + 26)^k = \binom{k}{0}

Vedclass · Accepted Answer

$676$

Vedclass · Answer

(A) Let $f(k) = 3^{3k} - 26k - 1 = 27^k - 26k - 1$. Using the Binomial Theorem,we can write $27^k$ as $(1 + 26)^k$. $27^k = (1 + 26)^k = \binom{k}{0} + \binom{k}{1}(26) + \binom{k}{2}(26)^2 + \dots + \binom{k}{k}(26)^k$. $27^k = 1 + 26k + \binom{k}{2}(26)^2 + \dots + (26)^k$. Substituting this into the expression for $f(k)$: $f(k) = (1 + 26k + \binom{k}{2}(26)^2 + \dots + (26)^k) - 26k - 1$. $f(k) = \binom{k}{2}(26)^2 + \binom{k}{3}(26)^3 + \dots + (26)^k$. All terms in this expansion are divisible by $(26)^2 = 676$. Therefore,$3^{3k} - 26k - 1$ is divisible by $676$.

If $k \in N$,then $3^{3k} - 26k - 1$ is divisible by

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