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If f(x)=(1+x)^2, then the value of f(x0)...

If `f(x)=(1+x)^2,` then the value of `f(x0)+f^(prime)(0)` `+(f^(0))/(2!)+(f^(0))/(3!)+(f^n(0))/(n !)dot`

Text Solution

Verified by Experts

The correct Answer is:
`2^(n)`
`f(x)=(1+x)^(n),f(0)=1`
`therefore" "f'(x)=n(1+x)^(n-1),f'(0)=n`
`therefore" "f''(x)=n(n-1)(1+x)^(n-2),f''(0)=n(n-1)`
Similarly, proceeding, we have
`f'''(0)=n(n-1)(n-2)`
`f''''(0)=n(n-1)(n-2)(n-3)` and so on.
`f^(n)(0)=n(n-1)(n-2)(n-3)...1`
`"or "f(0)+f'(0)+(f''(0))/(2!)+(f'''(0))/(3!)+...(f^(n)(0))/(n!)+`
`=1+n+(n(n-1))/(2!)+(n(n-1)(n-2))/(3!)+...+(n(n-1)(n-2)...1)/(n!)+...`
`=""^(n)C_(0)+""^(n)C_(1)+""^(n)C_(2)...+""^(n)C_(n-1)+""^(n)C_(n)=2^(n)`
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