The expression frac{k^5}{5} + frac{k^3}{3} + | vedclass

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

Question

(A) Let $P(k) = \frac{k^5}{5} + \frac{k^3}{3} + \frac{7k}{15}$ where $k \in N$. We can rewrite the expression as: $P(k) = \frac{3k^5 + 5k^3 + 7k}{15}

Vedclass · Accepted Answer

$A$ natural number

Vedclass · Answer

(A) Let $P(k) = \frac{k^5}{5} + \frac{k^3}{3} + \frac{7k}{15}$ where $k \in N$. We can rewrite the expression as: $P(k) = \frac{3k^5 + 5k^3 + 7k}{15} = \frac{3k^5 + 5k^3 - 8k + 15k}{15} = \frac{3(k^5 - k) + 5(k^3 - k) + 15k}{15}$ $P(k) = \frac{3}{15}(k^5 - k) + \frac{5}{15}(k^3 - k) + \frac{15k}{15}$ $P(k) = \frac{1}{5}(k^5 - k) + \frac{1}{3}(k^3 - k) + k$ Since $(k^5 - k)$ is divisible by $5$ and $(k^3 - k)$ is divisible by $3$ for all $k \in N$ (by Fermat's Little Theorem),the expression results in an integer. Since $k \in N$,the sum is always a natural number.

The expression $\frac{k^5}{5} + \frac{k^3}{3} + \frac{7k}{15}$ is always which of the following for $k \in N$?

Similar Questions

$(2^{3n} - 1)$ will be divisible by $(\forall n \in N)$

$(13)^{507}$ when divided by $9$ leaves the remainder :-

If $3^{2n+2}-8n-9$ is divisible by $2^p$ for all $n \in N$,then the maximum value of $p$ is

For any natural number $n$,$(15 \times 5^{2n}) + (2 \times 2^{3n})$ is divisible by

$\sqrt {\underbrace {111........1}_{200\,\text{digits}} - \underbrace {222.......2}_{100\,\text{digits}}} $ equals :-

Vedclass Test Series

Exam Paper Generator

Online Exam Module