A) 950
B) 975
C) 990
D) 1010
Correct Answer: C
Solution :
| [c] Let |
| \[\begin{matrix} {{S}_{1}}=2+3+5+9+16+........+{{x}_{n}} \\ {{S}_{1}}=2+3+5+9+........\,{{x}_{n-1}}+{{x}_{n}} \\ \begin{align} & \_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_ \\ & O=2+[1+2+4+7+....to\,\,(n-1)\,terms]-{{x}_{n}} \\ \end{align} \\ \end{matrix}\] |
| \[\therefore \,\,\,{{x}_{n}}=2+[1+2+4+7+......to\,(n-1)\,terms]\] |
| Again let |
| \[\,\,\,\begin{matrix} {{S}_{2}}=1+2+4+7+.........+{{t}_{n-1}} \\ {{S}_{2}}=1+2+4+7........+{{t}_{n-2}}+{{t}_{n-1}} \\ \begin{align} & \_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_ \\ & O=1+[1+2+3+.....+(n-2)\,\,terms]-{{t}_{n-1}} \\ \end{align} \\ \end{matrix}\] |
| \[{{t}_{n-1}}=1+\frac{(n-2)(n-1)}{2}=\frac{{{n}^{2}}-3n+4}{2}\] |
| \[\therefore \,\,\,{{S}_{2}}=\sum\limits_{n=1}^{n-1}{{{t}_{n-1}}=\frac{1}{2}\Sigma {{n}^{2}}-\frac{3}{2}\Sigma n+2\Sigma 1}\] |
| \[=\frac{1}{2}\frac{(n-1)\,\,n(2n-1)}{6}-\frac{3}{2}\frac{n(n-1)}{2}+2(n-1)\] |
| \[=(n-1)\,\left[ \frac{2{{n}^{2}}-n}{12}-\frac{3n}{4}+2 \right]\] |
| \[=\frac{n-1}{12}\left[ 2{{n}^{2}}-n-9n+24 \right]\] |
| \[=\frac{n-1}{6}\left[ {{n}^{2}}-5n+12 \right]=\frac{{{n}^{3}}-6{{n}^{2}}+17n-12}{6}\] |
| \[\therefore \,\,\,{{x}_{n}}=2+{{S}_{2}}=2+\frac{{{n}^{3}}-6{{n}^{2}}+17n-12}{6}\] |
| \[=\frac{{{n}^{3}}-6{{n}^{2}}+17n}{6}\] |
| So, \[{{x}_{20}}=990\] |
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