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Using the principle of mathematical induction, prove each of the following for all n ϵ N:

12 + 32 + 52 + 72 +.......+ (2n - 1)2\(\frac{n(2n-1)(2n+1)}3\)

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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

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