Question

Prove, by mathematical Induction, that
(n+1)^2+(n+2)^2 + (n+3)^2 ... + (2n)^2 =(n(2n+1)(7n+1))/6
is true
for
all natural numbers n.​

Answer 1

Step-by-step explanation:

step 1 we shall prove result is true for n=1

L.H.S=(2.1)²

=4

R.H.S=(1(2+1)(7+1)/6

=24/6

=4

result is true for n=1

step 2 let us assume result is true for n=1

(n+1)^2+(n+2)^2+(n+3)^2 + (2k)^2=(k(2k+1) 1

step 3 we shall now prove result is true for n=k+1

(n+1)^2+(n+2)^2+(n+3)^2 + (2(k+1)²=(k+1(2(k+1)+1)

l.h.s. using 1

2k^2+2(k^2+1^2+2k)

2k^2+2k^2+2+4k

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

R.H.S=(k+1(2k+3)

(k+1)(2k+3)

hope it helps you follow me also mark my answer as brainliest