Find the sum of the following series up to n terms:

(i) 5 + 55 + 555 + …

(ii) .6 + .66 + . 666 + …

Asked by Pragya Singh | 1 year ago |  98

Solution :-

(i) 5 + 55 + 555 + …

Let Sn = 5 + 55 + 555 + ….. up to n terms

$$\dfrac{5}{9}[9+99+999+....to\;n \;terms]$$

$$\dfrac{5}{9}[(10-1)+(10^2-1)+(10^3-1)$$

$$+....to\;n \;terms]$$

$$\dfrac{5}{9}[(10+10^2+10^3+....to\;n\;terms)$$

$$-(1+1+to\;n\;terms)]$$

$$\dfrac{5}{9}[\dfrac{10(10^n-1)}{10-1}-n]$$

$$\dfrac{5}{9}[\dfrac{10(10^n-1)}{9}-n]$$

$$\dfrac{50}{81}(10^n-1)-\dfrac{5n}{9}$$

Therefore, the sum of n terms of the given series is

$$\dfrac{50}{81}(10^n-1)-\dfrac{5n}{9}$$

(ii) Given, .6 + .66 + . 666 + …

Let Sn = 06. + 0.66 + 0.666 + … up to n term

$$\dfrac{6}{9}[(1-\dfrac{1}{10})+(1-\dfrac{1}{10^2})+(1-\dfrac{1}{10^3})+....to\;n\;terms]$$

$$\dfrac{2}{3}[(1+1+...to\;n\;terms)-$$

$$\dfrac{1}{10}(1+\dfrac{1}{10}+\dfrac{1}{10^2}to\;n\;terms)]$$

$$=\dfrac{2}{3}[n-10(1-\dfrac{(\dfrac{1}{10})^n}{1-(\dfrac{1}{10})})]$$

$$\dfrac{2}{3}n-\dfrac{2}{30}\times \dfrac{10}{9}(1-10^{-n})$$

$$\dfrac{2}{3}n-\dfrac{2}{27}(1-10^{-n})$$

Therefore, the sum of n terms of the given series is

$$\dfrac{2}{3}n-\dfrac{2}{27}(1-10^{-n})$$

Answered by Abhisek | 1 year ago

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