Sequence And Series Question 351
Question: Let $ S_{n}(1\le n\le 9) $ denotes the sum of n terms of series 1+22+333+……………..+9999999999, then for $ 2\le n\le 9 $
Options:
A) $ S_{n}-{S_{n-1}}=\frac{1}{9}(10^{n}-n^{2}+n) $
B) $ S_{n}=\frac{1}{9}(10^{n}-n^{2}+2n-2) $
C) $ 9,(S_{n}-{S_{n-1}})=n,(10^{n}-1) $
D) None of these
Show Answer
Answer:
Correct Answer: C
Solution:
[c] $ S_{n}=\frac{1}{9}(9)+\frac{2}{9}(99)+\frac{3}{9}(999)+… $ $ =\frac{1}{9}[ 10+{{2.10}^{2}}+{{3.10}^{3}}+… ]-\frac{1}{9}[1+2+3+…] $ $ =\frac{1}{9}S-\frac{1}{9}\frac{n(n+1)}{2} $ $ S=10+{{2.10}^{2}}+{{3.10}^{3}}+….+n10^{n} $ $ \frac{10,S=10^{2}+{{2.10}^{3}}+….+(n-1)10^{n}+n.{10^{n+1}}}{-9s=(10+10^{2}+….+10^{n})-n{10^{n+1}}} $ $ S=\frac{n}{9}{10^{n+1}}-\frac{{10^{n+1}}-10}{81} $
$ \therefore S_{n}=\frac{n}{81}{10^{n+1}}-\frac{{10^{n+1}}-10}{9.81}-\frac{1}{9}\frac{n(n+1)}{2} $
$ \therefore 9S_{n}=\frac{(9n-1){10^{n+1}}}{81}+\frac{10}{81}-\frac{n(n+1)}{2} $
$ \therefore 9(S_{n}-{S_{n-1}})=\frac{10^{n}}{81} $ $ { 10(9n-1)-(9n-10) }-n=n(10^{n}-1) $
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