# Prove by method of induction, for all n ∈ N: 11.3+13.5+15.7+...+1(2n-1)(2n+1)=n2n+1 - Mathematics and Statistics

Sum

Prove by method of induction, for all n ∈ N:

1/(1.3) + 1/(3.5) + 1/(5.7) + ... + 1/((2"n" - 1)(2"n" + 1)) = "n"/(2"n" + 1)

#### Solution

Let P(n) ≡ 1/(1.3) + 1/(3.5) + 1/(5.7) + ... + 1/((2"n" - 1)(2"n" + 1)) = "n"/(2"n" + 1)

Step 1:

For n = 1

L.H.S. = 1/(1.3) = 1/3

R.H.S. = 1/(2(1) + 1) = 1/3

∴ L.H.S. = R.H.S. for n = 1.

∴ P(1) is true.

Step 2:

Let us assume that for some k ∈ N, P(k) is true,

i.e., 1/(1.3) + 1/(3.5) + 1/(5.7) + ... + 1/((2"k" - 1)(2"k" + 1)) = "k"/(2"k" + 1)   ...(1)

Step 3:

To prove that P(k + 1) is true, i.e., to prove that

1/(1.3) + 1/(3.5) + 1/(5.7) + ... + 1/((2"k" - 1)(2"k" + 1)) + 1/((2"k" + 1)(2"k" + 3)) = ("k" + 1)/(2"k" + 3)

Now, L.H.S.

= 1/(1.3) + 1/(3.5) + 1/(5.7) + ... + 1/((2"k" - 1)(2"k" + 1)) + 1/((2"k" + 1)(2"k" + 3))

= "k"/(2"k" + 1) + 1/((2"k" + 1)(2"k" + 3))  ...[By (1)]

= 1/(2"k" + 1)["k" + 1/(2"k" + 3)]

= 1/(2"k" + 1)[(2"k"^2 + 3"k" + 1)/(2"k" + 3)]

= 1/((2"k" + 1)).(("k" + 1)(2"k" + 1))/(2"k" + 3)

= ("k" + 1)/(2"k" + 3)

= R.H.S.

∴ P(k + 1) is true.

Step 4:

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

i.e., 1/(1.3) + 1/(3.5) + 1/(5.7) + ... + 1/((2"n" - 1)(2"n" + 1)) = "n"/(2"n" + 1), for all n ∈ N.

Concept: Principle of Mathematical Induction
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#### APPEARS IN

Balbharati Mathematics and Statistics 2 (Arts and Science) 11th Standard Maharashtra State Board
Chapter 4 Methods of Induction and Binomial Theorem
Exercise 4.1 | Q 8 | Page 73