Question

Prove by method of induction; for all $n\in N$ $3+7+11+\dots ...+$ to $n$ terms $=n(2n+1)$

Answer 1

$3+7+11+15+...\dots \dots ..n$ terms $=n(2n+1)$

Put $n=1$

LHS$=3$

RHS$=1\left(2\right(1)+1)=2+1=3$

LHS$=$RHS

Let us assume that the result is true for $n=k$

$3+7+11+15+...\dots .+(4k-1)=k(2k+1)$

Let us prove this result for $n=k+1$

$3+7+11+15+...\dots \dots +(4k-1)+(4k+3)=(k+1)(2k+3)$

LHS$=3+7+11+...\dots \dots \dots +(4k-1)+(4k+3)$

$=k(2k+1)+(4k+3)$ (By assumption for k)

$=2k^2+k+4k+3$

$=2k^2+3k+2k+3$

$=k(2k+3)+1(2k+3)=(k+1)(2k+3)$

$=$RHS.

So, the result is true for $n=k+1$

Hence by induction principle, the given result is true for all natural numbers.