Question -
Answer -
Let P (n): 12 + 32 + 52 +… + (2n – 1)2 = 1/3 n (4n2 – 1)
Let us check for n = 1,
P (1): (2.1 – 1)2 = 1/3 × 1 × (4 – 1)
: 1 = 1
P (n) is true for n = 1.
Now, let us check for P (n) is true for n = k, and have toprove that P (k + 1) is true.
P (k): 12 + 32 + 52 +… + (2k – 1)2 = 1/3 k (4k2 – 1) … (i)
So,
12 + 32 + 52 +… + (2k – 1)2 + (2k + 1)2
Now, substituting the value of P (k) we get,
= 1/3 k (4k2 – 1) + (2k + 1)2 byusing equation (i)
= 1/3 k (2k + 1) (2k – 1) + (2k + 1)2
= (2k + 1) [{k(2k-1)/3} + (2k+1)]
= (2k + 1) [2k2 – k + 3(2k+1)] / 3
= (2k + 1) [2k2 – k + 6k + 3] / 3
= [(2k+1) 2k2 + 5k + 3] /3
= [(2k+1) (2k(k+1)) + 3 (k+1)] /3
= [(2k+1) (2k+3) (k+1)] /3
= (k+1)/3 [4k2 + 6k + 2k + 3]
= (k+1)/3 [4k2 + 8k – 1]
= (k+1)/3 [4(k+1)2 – 1]
P (n) is true for n = k + 1
Hence, P (n) is true for all n ∈ N.