A) \[100\]
B) \[90\]
C) \[900\]
D) \[99\]
Correct Answer: A
Solution :
Given, series \[1+3+7+13+21+.....\] Also, \[{{t}_{n}}=9901\] …..(i) Let \[{{S}_{n}}=1+3+7+12+21+.......n\,terms\] and \[{{S}_{n}}=1+3+7+13+....n\,\,terms\] On subtracting \[0=(1+2+4+6+8+....)-{{t}_{n}}\] \[{{t}_{n}}=1+2+4+6+8+......n\,\,terms\] \[{{t}_{n}}=1+2[1+2+3+4+.......(n-1)terms]\] \[{{t}_{n}}=1+2\left[ \frac{(n-1)\,(n-1+1)}{2} \right]\] \[{{t}_{n}}=1+n(n-1)\] \[9901=1+n(n-1)\] [From Eq. (i)] \[{{n}^{2}}-n-9900=0\] \[{{n}^{2}}-100n+99n-9900=0\] \[n(n-100)+99(n-100)=0\] \[(n-100)(n+99)=0\] \[\Rightarrow \] \[n=100\] \[(n=-99,\,neglecting)\] (because terms not negative)
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