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Question
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.
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