To Prove:
12 + 32 + 52 + 72 +.......+ (2n - 1)2 = \(\frac{n(2n-1)(2n+1)}3\)
Steps to prove by mathematical induction:
Let P(n) be a statement involving the natural number n such that
(i) P(1) is true
(ii) P(k + 1) is true, whenever P(k) is true
Then P(n) is true for all n ϵ N
Therefore,
Let P(n): 12 + 32 + 52 + 72 +.......+ (2n - 1)2 = \(\frac{n(2n-1)(2n+1)}3\)
Step 1:
P(1) = \(\frac{1(2-1)(2+1)}3\) = \(\frac{3}3\) = 1
Therefore, P(1) is true
Step 2:
Let P(k) is true Then,
P(k): 12 + 32 + 52 + 72 + … + (2k – 1)2 = \(\frac{k(2k-1)(2kj1)}3\)
Now,
12 + 32 + 52 + 72 + … + (2(k + 1)–1)2 = \(\frac{k(2k-1)(2k+1)}3\) + (2k + 2 - 1)2
(Splitting the middle term)
= \(\frac{(k+1)(2k+1)(2k+3)}3\)
= P(k + 1)
Hence, P(k + 1) is true whenever P(k) is true
Hence, by the principle of mathematical induction, we have
12 + 32 + 52 + 72 + … + (2n – 1)2 = \(\frac{n(2n-1)(2n+1)}3\)
for all n ϵ N