When P is a natural number,then {P^{n + 1}} + | vedclass

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

Question

(C) For $n = 1$,we get,${P^{1 + 1}} + {(P + 1)^{2(1) - 1}} = {P^2} + {(P + 1)^1} = {P^2} + P + 1$. This expression is clearly divisible by ${P^2} + P

Vedclass · Accepted Answer

${P^2} + P + 1$

Vedclass · Answer

(C) For $n = 1$,we get,${P^{1 + 1}} + {(P + 1)^{2(1) - 1}} = {P^2} + {(P + 1)^1} = {P^2} + P + 1$. This expression is clearly divisible by ${P^2} + P + 1$. Let us assume that the given result is true for $n = m \in N$,i.e.,${P^{m + 1}} + {(P + 1)^{2m - 1}} = k({P^2} + P + 1)$ for some $k \in N$ .....$(i)$ Now,for $n = m + 1$: ${P^{(m + 1) + 1}} + {(P + 1)^{2(m + 1) - 1}} = {P^{m + 2}} + {(P + 1)^{2m + 1}} = {P^{m + 2}} + {(P + 1)^2}{(P + 1)^{2m - 1}}$. Using equation $(i)$,we substitute ${(P + 1)^{2m - 1}} = k({P^2} + P + 1) - {P^{m + 1}}$: $= {P^{m + 2}} + {(P + 1)^2}[k({P^2} + P + 1) - {P^{m + 1}}] = {P^{m + 2}} + {(P + 1)^2} \cdot k({P^2} + P + 1) - {(P + 1)^2}{P^{m + 1}} = {P^{m + 1}}[P - {(P + 1)^2}] + {(P + 1)^2} \cdot k({P^2} + P + 1) = {P^{m + 1}}[P - ({P^2} + 2P + 1)] + {(P + 1)^2} \cdot k({P^2} + P + 1) = {P^{m + 1}}[-{P^2} - P - 1] + {(P + 1)^2} \cdot k({P^2} + P + 1) = -{P^{m + 1}}({P^2} + P + 1) + {(P + 1)^2} \cdot k({P^2} + P + 1) = ({P^2} + P + 1)[k{(P + 1)^2} - {P^{m + 1}}]$. Since this is a multiple of ${P^2} + P + 1$,the result is true for $n = m + 1$. By the principle of mathematical induction,the result is true for all $n \in N$.

When $P$ is a natural number,then ${P^{n + 1}} + {(P + 1)^{2n - 1}}$ is divisible by

Similar Questions

Prove the following by using the principle of mathematical induction for all $n \in N$:
$\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\ldots+\frac{1}{2^{n}}=1-\frac{1}{2^{n}}$

Prove the following by using the principle of mathematical induction for all $n \in N$:
$1 \cdot 3 + 3 \cdot 5 + 5 \cdot 7 + \ldots + (2n - 1)(2n + 1) = \frac{n(4n^2 + 6n - 1)}{3}$

Prove the statement by the Principle of Mathematical Induction:
$2n < (n+2)!$ for all natural numbers $n$.

Prove the statement by the Principle of Mathematical Induction: $n(n^{2}+5)$ is divisible by $6$,for each natural number $n$.

Prove the following by using the principle of mathematical induction for all $n \in N:$
$4+8+12+\ldots+4n = 2n(n+1)$

Vedclass Test Series

Exam Paper Generator

Online Exam Module