​ Prove by the method of induction, for all n ∈ N. 1/3.5 + 1/5.7 + 1/7.9+....to n terms = ​n/3(2n + 3)

Question

Prove by the method of induction, for all n ∈ N.

1/3.5 + 1/5.7 + 1/7.9+....to n terms = \(\frac{n}{3(2n+3)}\)

Answer 1

Let P(n) =  1/3.5 + 1/5.7 + 1/7.9+....to n terms = \(\frac{n}{3(2n+3)}\), for all n ∈ N.

But first factor in each term of the denominator,

i.e., 3, 5, 7, ….. are in A.P. with a = 3 and d = 2.

∴ nth term = a + (n – 1)d = 3 + (n – 1) 2 = (2n + 1)

Also, second factor in each term of the denominator,

i.e., 5, 7, 9, … are in A.P. with a = 5 and d = 2.

∴ nth term = a + (n – 1) d = 5 + (n – 1) 2 = (2n + 3)

∴ nth term, t_n = \(\frac{1}{(2n+1)(2n+3)}\)

P(n) = 1/3.5 + 1/5.7 + 1/7.9 + ....+ \(\frac{1}{(2n+1)(2n+3)}\)

 = \(\frac{n}{3(2n+3)}\) 

Step I:

Put n = 1

L.H.S. = 1/3.5 = 1/15

R.H.S = \(\frac{1}{2[2(1)+3]}=\frac1{3(2+3)}=\frac1{15}\)

∴ L.H.S. = R.H.S.

∴ P(n) is true for n = 1.

Step II: Let us assume that P(n) is true for n = k.

  1/3.5 + 1/5.7 + 1/7.9 + ....+ \(\frac{1}{(2k+1)(2k+3)}\) =  \(\frac{k}{3(2k+3)}\).....(i)

Step III: We have to prove that P(n) is true for n = k + 1,

i.e., to prove that

∴ P(n) is true for n = k + 1.

Step IV:

From all the steps above, by the principle of mathematical induction, P(n) is true for all n ∈ N.

\(\therefore\) 

1/3.5 + 1/5.7 + 1/7.9 + ....to n terms  = \(\frac{n}{3(2n+3)}\), for all n ∈ N.