1.

`(1)/(n!)+(1)/(2!(n-2)!)+(1)/(4!(n-4)!)`+….to n terms is equal toA. `(2n-1)/(2!)`B. `(2^(n))/(n+1)!`C. `(2^(n))/(n!)`D. `(2^(n)-2)/(n-1)!`

Answer» Answer:
We have
`(1)n!+(1)/(2!(n-2)!)+(1)/(4!(n-4)!)`+…
`=(1)/(n!){(1+.^(n)C_(2)+.^(n)C_(4)+…}`
`=(1)/(n!){.^(n)C_(0)+.^(n)C_(2)+.^(n)C_(4)+…}=(1)/(n!)(2^(n-1))=(2^(n-1))/(n!)`


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