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Prove by method of induction, for all n ∈ N: 13.5+15.7+17.9+... to n terms = n3(2n+3)

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Question

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

`1/(3.5) + 1/(5.7) + 1/(7.9) + ...` to n terms = `"n"/(3(2"n" + 3))`

Sum
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Solution

Let P(n) ≡ `1/(3.5) + 1/(5.7) + 1/(7.9) + ...` to n terms = `"n"/(3(2"n" + 3))`, for all n ∈ N

But the 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 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 tn = `1/((2"n" + 1)(2"n" + 3))`

∴ P(n) ≡ `1/(3.5) + 1/(5.7) + 1/(7.9) + ...  + 1/((2"n" + 1)(2"n" + 3)) = "n"/(3(2"n" + 3))`

Step I:

Put n = 1

L.H.S. = `1/(3.5) = 1/15`

R.H.S. = `1/(3[2(1) + 3]) = 1/(3(2 + 3)) = 1/15` = L.H.S.

∴ P(n) is true for n = 1

Step II:

Let us consider that P(n) is true for n = k

∴  `1/(3.5) + 1/(5.7) + 1/(7.9) + ...  + 1/((2"k" + 1).(2"k" + 3)) = "k"/(3(2"k" + 3))`   ...(i)

Step III:

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

i.e., to prove that

`1/(3.5) + 1/(5.7) + 1/(7.9) + ...  + 1/((2"k" + 3).(2"k" + 5)) = ("k" + 1)/(3(2"k" + 5))` 

L.H.S. = `1/(3.5) + 1/(5.7) + 1/(7.9) + ...  + 1/((2"k" + 3)(2"k" + 5))` 

= `1/(3.5) + 1/(5.7) + 1/(7.9) + ...  + 1/((2"k" + 1)(2"k" + 3)) + 1/((2"k" + 3)(2"k" + 5))`

= `"k"/(3(2"k" + 3)) + 1/((2"k" + 3)(2"k" + 5))` ...[From (i)]

= `("k"(2"k" + 5) + 3)/(3(2"k" + 3)(2"k" + 5))`

= `(2"k"^2 + 5"k" + 3)/(3(2"k" + 3)(2"k" + 5))`

= `(2"k"^2 + 2"k" + 3"k" + 3)/(3(2"k" + 3)(2"k" + 5)`

= `(2"k"("k" + 1) + 3("k" + 1))/(3(2"k" + 3)(2"k" + 5))`

= `((2"k" + 3)("k" + 1))/(3("2k" + 3)(2"k" + 5))`

= `("k" + 1)/(3(2"k" + 5))`

= R.H.S.

∴ P(n) is true for n = k + 1

Step IV:

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

∴ `1/(3.5) + 1/(5.7) + 1/(7.9) + ...` to n terms = `"n"/(3(2"n" + 3))` for all n ∈ N.

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Chapter 4: Methods of Induction and Binomial Theorem - Exercise 4.1 [Page 74]

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Balbharati Mathematics and Statistics (Arts and Science) Part 2 [English] Standard 11 Maharashtra State Board
Chapter 4 Methods of Induction and Binomial Theorem
Exercise 4.1 | Q 9 | Page 74

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