sigma k=1 to n (2k-1)whole square = n(2n-1)(2n+1)/3 - Maths - Principle of Mathematical Induction

Question

sigma k=1 to n (2k-1)whole square = n(2n-1)(2n+1)/3

Ajanta Trivedi · Accepted Answer

to prove the above result by
mathematical induction method, we will follow three
steps:
step1: let us prove the result for
n=1
∑k=11 (2k-1)2=1 RHS=1*(2-1)*(2+1)3=1
thus the result is true for n =1
step 2:
let us assume that result holds for n = p
therefore∑k=1p(2k-1)2=p(2p-1)(2p+1)3  (1)
step3:
now with the help of assumption, we will prove that result is true
for n = p+1
∑k=1p+1 (2k-1)2=∑k=1p (2k-1)2+[2(p+1)-1]2 =p(2p-1)(2p+1)3+2p+12=(2p+1)
p(2p-1)3+(2p+1)=(2p+1)3 2p2-p+6p+3=(2p+1)3 2p2+5p+3=(2p+1)3
2p2+2p+3p+3
=(2p+1)3 2p(p+1)+3(p+1)=(2p+1)3 (p+1)(2p+3)=(p+1) [2(p+1)+1]
[2(p+1)-1]3

thus the result holds for n = p+1
therefore result is true for all integer values of n

hope this helps you