Find the sum upto n terms of following : ■ 5+55+555+.... ■0.3+0.33+0.333+..... ■7+77+777+....

Question

This is from the chapter geometric progression.

Answer 1

???????? geometrik progresshun ???????

0.3+0.33+0.333+...=0.6666666...3

number av "6"=n-1

Answer 2

Each series is based on a common series, for example: S=1+11+111+1111+... So if we work out Sn for two of the series we only need to multiply by 5 and 7 respectively to get the sum to n terms of each.

We can see that S consists of the sum of several GPs: 1+10+100+... is a GP with sum (10^n-1)/9, where n starts at 1. The next series is (10^(n-1)-1)/9. If, for example, we take n=2, we get (10^2-1)/9+(10-1)/9=11+1. For n=5, we get (10^5-1)/9+(10^4-1)/9+(10^3-1)/9+(10^2-1)/9+(10-1)/9=11111+1111+111+11+1.

What we have here is 10^5/9+10^4/9+...+10/9-5/9. This is a GP with a=10/9 and r=10 less 5/9. The sum of the GP by itself is (10/9)(10^5-1)/9=10(10^5-1)/81=11110/9. Subtract 5/9 from this: 12345. The general expression is 10(10^n-1)/81-n/9. For the first and third series then, the sum to n terms is 50(10^n-1)/81-5n/9 and 70(10^n-1)/81-7n/9.

The middle series is slightly different in that r=0.1 instead of 10.

S=3(0.1+0.11+...)=3((0.1)+(0.1+0.01)+(0.1+0.01+0.001)+...). Here a=0.1, r=0.1 for each sub-series.

As in the earlier exercise, take n=5: the sum of the main series is 0.1((1-0.1)/0.9+0.1(1-0.1^2)/0.9+0.1(1-0.1^3)0.9+0.1(1-0.1^4)/0.9+0.1(1-0.1^5)/0.9=0.1((1-0.1)/0.9+...+(1-0.1^5)/0.9)=0.1(5/0.9-(0.1^5)/0.9-(0.1^4)/0.9-...-(0.1)/0.9)=0.1(5/0.9-S'), where S' is the GP with a=0.1/0.9=1/9 and r=0.1: 0.1/0.9+0.1^2/0.9+0.1^3/0.9+0.1^3/0.9+0.1^4/0.9=(1/9)(1+0.1+...+0.1^(5-1))

or for n: (1/9)(1+0.1+...+0.1^(n-1))=(1/9)(1-0.1^n)/(1-0.1)=(1-0.1^n)/8.1.

So the complete series is 0.1(n/0.9-(1-0.1^n)/8.1). S is therefore 0.3(n/0.9-(1-0.1^n)/8.1).

(When n is large, this approximates to n/3-1/27 or (9n-1)/27.)

[Apologies for the length of time to produce the complete solution.]

Answer 3

5+55+555+... to n terms

nth term
= 5 + 50 + 500 + ... + 5x10^(n-1)
= 5 * [10^n - 1]/(10 - 1)
= (5/9)[10^n - 1]

Σ (5/9)[10^n - 1]
= (5/9) [Σ10^n - Σ1]
= (5/9) [10 * (10^n - 1)/(10 - 1) - n]
= (5/9)[(10/9)*10^n - (10/9) - n]
= (50/81) * (10^n - 1) - 5n/9.

7, 77, 777, 7777 ….to n terms.
Sn=7+77+777+7777+……+to n terms

Introducing 9
= (7/9) [9+99+999+……+ to n terms]
= (7/9) [(10-1) + (100-1) + (1000-1) +……...+ to n term]
= (7/9) [10+100+1000+……... n terms – (1+1+1+1+…… n terms)
= (7/9) [10+100+1000+………n terms –n]

Here a=10, r=10
= (7/9) [10(10n-1)/ (9) - n]

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