Let the given statement be P(n), i.e.,
P (n) : (1+

3

1

)(1+

5

4

)(1+

7

9

) ... (1+

(2n+1)

n2

) = (n+1)2
First we prove that the statement is true for n = 1.
P (1) : (1+

3

1

) = (1+1)2 ⇒ 1 + 3 = 22 ⇒ 4 = 4 , which is true
Assume P(k) is true for some positive integer k, i.e.,
(1+

3

1

)(1+

5

4

)(1+

7

9

) ... (1+

(2k+1)

k2

) = (k+1)2
... (i)
Now we shall prove that P(k + 1) is also true.
For this we have to prove that
(1+

3

1

)(1+

5

4

)(1+

7

9

) ... (1+

(2k+1)

k2

) (1+

2(k+1)+1

(k+1)2

) = [(k+1)+1]2
L.H.S. = (1+

3

1

)(1+

5

4

)(1+

7

9

) ... (1+

(2k+1)

k2

) (1+

2(k+1)+1

(k+1)2

)
= (k+1)2(1+

2(k+1)+1

(k+1)2

) [From (i)]
= (k+1)2

[(k+1)2+2(k+1)+1]

(k+1)2

= (k+1)2 + 2 (k + 1) + 1 = k2 + 2k + 1 = k2 + 2k + 1 + 2k + 2 + 1 = k2 + 4k + 4 = (k+2)2
= [(k+1)+1]2 = R.H.S.
Thus P(k + 1) is true, whenever P(k) is true.
Hence, by the principle of mathematical induction P(n) is true ∀ n ∈ N.