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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

Similar Questions

$a^n + b^n$ is divisible by which of the following if $n$ is any odd positive integer?

The expression $(1 + x)^n - nx - 1$ is divisible by (where $n \in N$ and $n > 1$):

The remainder when $3^{91}$ is divided by $80$ is:

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

Show that $9^{n+1}-8n-9$ is divisible by $64$ whenever $n$ is a positive integer.

Vedclass Test Series

Exam Paper Generator

Online Exam Module