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Question

If f(x)=xn, then find the value of f(1)+f1(1)1!+f2(1)2!+......+fn(1)n!, where fr(x) denotes the rth derivative of f(x) w.r.t.x

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Solution

f(x)=xnf(x)=1

f1(x)=nxn1f1(1)=n

f2(x)=n(n1)xn2f2(1)=n(n1)

f3(x)=n(n1)(n2)xn3f3(1)=n(n1)(n2)

fn(x)=n(n1)(n2).....[n(n1)]xn(na)

=n(n1)(n2).....1

fn(1)=n(n1)(n2).....1

f(1)+f(1)1!+f2(1)2!+.....+fn(1)n!

=1+n1!+n1!+n(n1)2!+....+n(n1)(n2)....1n!

=1+nC1+nC2+....+nCn

=nC0+nC1+nC2+.....+nCn

=2n

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