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Prove by method of induction, for all n ∈ N: 1.3 + 3.5 + 5.7 + ..... to n terms = n3(4n2+6n-1) - Mathematics and Statistics

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प्रश्न

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

1.3 + 3.5 + 5.7 + ..... to n terms = `"n"/3(4"n"^2 + 6"n" - 1)`

बेरीज
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उत्तर

Let P(n) ≡ 1.3 + 3.5 + 5.7 + ..... to n terms = `"n"/3(4"n"^2 + 6"n" - 1)`, for all n ∈ N

But the first factor in each term

i.e., 1, 3, 5 … are in A.P. with a = 1 and d = 2.

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

Also second factor in each term

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)

∴ nth term, tn = (2n – 1) (2n + 1)

∴ P(n) ≡ 1.3 + 3.5 + 5.7 + .... + (2n – 1) (2n + 1) = `"n"/3(4"n"^2 + 6"n" - 1)`

Step I:

Put n = 1

L.H.S. = 1.3 = 3

R.H.S. = `1/3[4(1)^2 + 6(1) - 1]` = 3 = 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 + 3.5 + 5.7 + ..... + (2k – 1)(2k + 1)

= `"k"/3(4"k"^2 + 6"k" - 1)`   ...(i)

Step III:

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

i.e., to prove that

1.3 + 3.5 + 5.7 + …. + [2(k + 1) – 1][2(k + 1) + 1]

= `(("k" + 1))/3[4("k" + 1)^2 + 6("k" + 1) - 1]`

= `(("k" + 1))/3(4"k"^2 + 8"k" + 4 + 6"k" + 6 - 1)`

= `(("k" + 1))/3(4"k"^2 + 14"k" + 9)`

L.H.S. = 1.3 + 3.5 + 5.7 + ... + [2(k + 1) – 1][2(k + 1) + 1]

= 1.3 + 3.5 + 5.7 + ... + (2k – 1)(2k + 1) + (2k + 1)(2k + 3)

= `"k"/3(4"k"^2 + 6"k" - 1) + (2"k" + 1) (2"k" + 3)`  ...[From (i)]

= `1/3[4"k"^3 + 6"k"^2 - "k" + 3(2"k" + 1)(2"k" + 3)]`

= `1/3(4"k"^3 + 6"k"^2 - "k" + 12"k"^2 + 24"k" + 9)`

= `1/3(4"k"^3 + 18"k"^2 + 23"k" + 9)`

= `1/3("k" + 1)(4"k"^2 + 14"k" + 9)`

= 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 + 3.5 + 5.7 + ..... to n terms = `"n"/3(4"n"^2 + 6"n" - 1)` for all n ∈ N.

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पाठ 4: Methods of Induction and Binomial Theorem - Exercise 4.1 [पृष्ठ ७३]

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बालभारती Mathematics and Statistics 2 (Arts and Science) [English] Standard 11 Maharashtra State Board
पाठ 4 Methods of Induction and Binomial Theorem
Exercise 4.1 | Q 7 | पृष्ठ ७३

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