prove that the sum to n terms of the series 11+103+1005+ - Maths - Sequences and Series

Question

prove that the sum to n terms of the series 11+103+1005+ is 10 /
9 (10to the power n - 1)+n (sq)

Santosh Kumar · Accepted Answer

S = 11 + 103 + 1005 + â‡’ S = (10 + 1) + (100 + 2) + (1000 + 5) + $\Rightarrow S = (10^{1} + 1) + (10^{2} + 3) + (10^{3} + 5) + \Rightarrow S = (10^{1} + 10^{2} + 10^{3} +) + (1 + 3 + 5 +) \Rightarrow S = S_{1} + S_{2} (1)$ Now, 101 + 102 + 103 + is a G P series Where, first term, a = 101 $Common ratio, r = \frac{10^{2}}{10} = 10^{2 - 1} = 10 S_{1} = \frac{a (r^{n} - 1)}{(r - 1)} ∴ 10^{1} + 10^{2} + 10^{3} + n^{} terms = \frac{10 (10^{n} - 1)}{10 - 1} = \frac{10 (10^{n} - 1)}{9}$ 1 + 3 + 5 + is an A P series Where, first term, a = 1 Common difference, d = 5 â€“ 3 = 2 $S_{2} = \frac{n}{2} [2 a + (n - 1) d] = \frac{n}{2} [2 (1) + (n - 1) 2] = \frac{n}{2} (2 + 2 n - 2) = n^{2} \Rightarrow S = \frac{10}{9} (10^{n} - 1) + n^{2} ∴ 11 + 103 + 1005 + n terms = \frac{10}{9} (10^{n} - 1) + n^{2}$

prove that the sum to n terms of the series 11+103+1005+.......... is 10 / 9 (10to the power n - 1)+n (sq).