Question -
Answer -
Let P (n): 1.3 + 3.5 + 5.7 + … + (2n – 1) (2n + 1) =n(4n2 + 6n – 1)/3
Let us check for n = 1,
P (1): (2.1 – 1) (2.1 + 1) = 1(4.12 + 6.1-1)/3
: 1×3 = 1(4+6-1)/3
: 3 = 3
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): 1.3 + 3.5 + 5.7 + … + (2k – 1) (2k + 1) = k(4k2 +6k – 1)/3 … (i)
So,
1.3 + 3.5 + 5.7 + … + (2k – 1) (2k + 1) + (2k + 1) (2k + 3)
Now, substituting the value of P (k) we get,
= k(4k2 + 6k – 1)/3 + (2k + 1) (2k + 3) byusing equation (i)
= [k(4k2 + 6k-1) + 3 (4k2 +6k + 2k + 3)] / 3
= [4k3 + 6k2 – k + 12k2 +18k + 6k + 9] /3
= [4k3 + 18k2 + 23k + 9] /3
= [4k3 + 4k2 + 14k2 +14k +9k + 9] /3
= [(k+1) (4k2 + 8k +4 + 6k + 6 – 1)] / 3
= [(k+1) 4[(k+1)2 + 6(k+1) -1]] /3
P (n) is true for n = k + 1
Hence, P (n) is true for all n ∈ N.