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`(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)!` |
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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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