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

Question

Answer the following:

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

8 + 17 + 26 + … + (9n – 1) = `"n"/2(9"n" + 7)`

Sum

Solution

Let P(n) ≡ 8 + 17 + 26 + … + (9n – 1) = `"n"/2(9"n" + 7)`, for all n ∈ N

Step I:

Put n = 1

L.H.S. = 8

R.H.S. = `1/2[9(1)+ 7]` = 8 = L.H.S.

∴ P(n) is true for n = 1

Step II:

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

∴ 8 + 17 + 26 + … + (9k – 1) = `"k"/2(9"k" + 7)` ...(i)

Step III:

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

i.e., 8 + 17 + 26 + … + [9(k + 1) – 1]

= `(("k" + 1))/2 [9("k" + 1) + 7]`

= `(("k" + 1))/2 (9"k" + 16)`

L.H.S. = 8 + 17 + 26 + … + [9(k + 1) – 1]

= 8 + 17 + 26 + … + (9k – 1) + [9(k + 1) – 1]

= `"k"/2 (9"k" + 7) + (9"k" + 8)` ...[From (i)]

= `(9"k"^2 + 7"k" + 18"k" + 16)/2`

= `(9"k"^2 + 25"k" + 16)/2`

= `(9"k"^2 + 9"k" + 16"k" + 16)/2`

= `(9"k"("k" + 1) + 16("k" + 1))/2`

= `(("k" + 1))/2(9"k" + 16)`

= 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.

∴ 8 + 17 + 26 + … + (9n – 1) = `"n"/2(9"n" + 7)` for all n ∈ N

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Principle of Mathematical Induction

Is there an error in this question or solution?

Q II. (1) (i)Q I. (10)Q II. (1) (ii)

Chapter 4: Methods of Induction and Binomial Theorem - Miscellaneous Exercise 4.2 [Page 85]

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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
Miscellaneous Exercise 4.2 | Q II. (1) (i) | Page 85

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