KCET2010MathematicsSequences and Series
The n th term of the series 1+3+7+13+21+ is 9901 . The value of n is
Options
- A100
- B90
- C900
- D99
Correct answer
A. 100
Step-by-step solution
Given, series 1+3+7+13+21+ Also, t _ n =9901 ...(i) Let S _ n =1+3+7+13+21+ n 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 [ ( n -1)( n -1+1) 2 ] t _ n =1+ n ( n -1)9901=1+n(n-1) [from Eq. (i) n ²- n -9900=0n²-100 n+99 n-9900=0 gathered n ( n -100)+99( n -100)=0 ( n -100)( n +99)=0 n =100 ( n =-99, neglecting ) gathered (because terms not negative)