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
Correct Answer: C
Solution :
| [c] \[{{S}_{n}}=\frac{1}{9}(9)+\frac{2}{9}(99)+\frac{3}{9}(999)+...\] |
| \[=\frac{1}{9}\left[ 10+{{2.10}^{2}}+{{3.10}^{3}}+... \right]-\frac{1}{9}[1+2+3+...]\] |
| \[=\frac{1}{9}S-\frac{1}{9}\frac{n(n+1)}{2}\] |
| \[S=10+{{2.10}^{2}}+{{3.10}^{3}}+....+n\,\,{{10}^{n}}\] |
| \[\frac{10\,S={{10}^{2}}+{{2.10}^{3}}+....+(n-1){{10}^{n}}+n.\,\,{{10}^{n+1}}}{-9s=(10+{{10}^{2}}+....+{{10}^{n}})-n\,\,{{10}^{n+1}}}\] |
| \[S=\frac{n}{9}{{10}^{n+1}}-\frac{{{10}^{n+1}}-10}{81}\] |
| \[\therefore {{S}_{n}}=\frac{n}{81}{{10}^{n+1}}-\frac{{{10}^{n+1}}-10}{9.81}-\frac{1}{9}\frac{n(n+1)}{2}\] |
| \[\therefore 9{{S}_{n}}=\frac{(9n-1){{10}^{n+1}}}{81}+\frac{10}{81}-\frac{n(n+1)}{2}\] |
| \[\therefore 9({{S}_{n}}-{{S}_{n-1}})=\frac{{{10}^{n}}}{81}\] |
| \[\left\{ 10(9n-1)-(9n-10) \right\}-n=n({{10}^{n}}-1)\] |
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