AP EAMCET2008MathematicsSequences and Series
_ k=1 ^ 1 k ! ( _ n=1 ^k 2^ n-1 ) is equal to
Options
- Ae
- Be^2+e
- Ce^2
- De^2-e
Correct answer
D. e^2-e
Step-by-step solution
aligned & _ k=1 ^ 1 k ! ( _ n=1 ^k 2^ n-1 ) &= _ k=1 ^ 1 k ! [1 (2^k-1 ) ] aligned aligned & = _ k=1 ^ 2^k-1 k ! & = _ k=1 ^ 2^k k ! - _ k=1 ^ 1 k ! & =e^2-1-(e-1) & =e^2-e aligned