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1/(n!)+1/(2!(n-2)!)+1/(4!(n-4)!)+ …… is ...

`1/(n!)+1/(2!(n-2)!)+1/(4!(n-4)!)+ ……` is equal to

Text Solution

Verified by Experts

The correct Answer is:
`(2^(n-1))/(n!)`
`(1)/(n!) + (1)/(2!(n-2)!) + (1)/(4!(n-4)!) + "……"`
`= 1/(n!) [.^(n)C_(0) + .^(n)C_(2) + .^(n)C_(4)+ "….." ] = (2^(n-1))/(n!)`
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