Maharashtra State BoardHSC Science (General) 11th
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Prove by method of induction, for all n ∈ N: 5 + 52 + 53 + .... + 5n = 54(5n-1) - Mathematics and Statistics

Sum

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

5 + 52 + 53 + .... + 5n = `5/4(5^"n" - 1)`

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Solution

Let P(n) ≡ 5 + 52 + 53 + .... + 5n = `5/4(5^"n" - 1)`, for all n ∈ N

Step I:

Put n = 1

L.H.S. = 5

R.H.S. = `5/4(5^1 - 1)` = 5 = L.H.S.

∴ P(n) is true for n = 1.

Step II:

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

∴ 5 + 52 + 53 + …. + 5k = `5/4(5^"k" - 1)`   ...(i)

Step III:

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

5 + 52 + 53 + …. + 5k+1 = `5/4(5^("k"+1) - 1)`

L.H.S. = 5 + 52 + 53 + …. + 5k+1 

= 5 + 52 + 53 + …. + 5k + 5k+1 

= `5/4(5^"k" - 1) + 5^("k"+1)`   ...[From (i)]

= `(5.5^"k" - 5 + 4.5^("k"+1))/4`

= `(5^("k"+1) + 4.5^("k"+1) - 5)/4`

= `(5.5^("k"+1) - 5)/4`

= `5/4(5^("k" + 1) - 1)`

= 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.

∴ 5 + 52 + 53 + .... + 5n = `5/4(5^"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 13 | Page 74
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