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Mathematics

Find the sum of the following to n terms: 7 + 77 + 777 + 7777 + ………

G.P.

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Answer

Sn=7+77+777+7777+ upto n terms=7(1+11+111+.....) upto n terms=79(9+99+999+.....) upto n terms=79[(101)+(1021)+(1031)+.......+(10n1)]=79[(10+102+103+.....+10n)(1+1+1+.....n times)]=79[(10+102+103+.....+10n)n] …….(1)S_n = 7 + 77 + 777 + 7777 + ……… \text{ upto n terms} \\[1em] = 7(1 + 11 + 111 +…..) \text{ upto n terms} \\[1em] = \dfrac{7}{9} (9 + 99 + 999 +…..) \text{ upto n terms} \\[1em] = \dfrac{7}{9}[(10 - 1) + (10^2 - 1) + (10^3 - 1)+…….+(10^n - 1)] \\[1em] = \dfrac{7}{9}[(10 + 10^2 + 10^3+…..+10^n) - (1 + 1 + 1+…..\text{n times})] \\[1em] = \dfrac{7}{9}[(10 + 10^2 + 10^3+…..+10^n) - n] \text{ …….(1)}

Now,

Calculating the sum of 10 + 102 + 103 + ……… + 10n.

a = 10

r = 10210\dfrac{10^2}{10} = 10

We know that,

The sum of the first n terms of a G.P. is given by:

Sn=a(rn1)r1S_n = \dfrac{a(r^n - 1)}{r - 1} [For r > 1]

Sn=10(10n1)101=10(10n1)9.\Rightarrow S_n = \dfrac{10(10^n - 1)}{10 - 1} \\[1em] = \dfrac{10(10^n - 1)}{9}.

Substitute and Simplify, the above value in equation (1), we get :

Sn=79[10(10n1)9n]=781[10×10n109n]=781[10n+19n10]S_n = \dfrac{7}{9}\Big[\dfrac{10(10^n - 1)}{9} - n\Big] \\[1em] = \dfrac{7}{81}\Big[10 \times 10^n - 10 - 9n\Big] \\[1em] = \dfrac{7}{81}\Big[10^{n + 1} - 9n - 10\Big]

Hence, Sn = 781[10n+19n10]\dfrac{7}{81}\Big[10^{n + 1} - 9n - 10\Big].

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