NIMCET 2007 — Mathematics PYQ
NIMCET | Mathematics | 2007If f(x)=xn, then f(1)+f′(1)1!+f″(1)2!+⋯+fn(1)n! is equal to:
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If f(x)=xn, then f(1)+f′(1)1!+f″(1)2!+⋯+fn(1)n! is equal to:
1. Identify the Derivatives:
Given the function f(x)=xn, let's find its successive derivatives:
f(x)=xn⟹f(1)=1n=1
f′(x)=nxn−1⟹f′(1)=n
f″(x)=n(n−1)xn−2⟹f″(1)=n(n−1)
f(k)(x)=n!(n−k)!xn−k⟹f(k)(1)=n!(n−k)!
2. Substitute into the Series:
The given expression is:
S=f(1)+f′(1)1!+f″(1)2!+⋯+fn(1)n!
Substituting the values we found:
S=1+n1!+n(n−1)2!+⋯+n!n!
3. Relate to Binomial Coefficients:
Notice that the terms are exactly the binomial coefficients nCr:
1=nC0
n1!=nC1
n(n−1)2!=nC2
n!n!=nCn
So, the expression becomes:
S=nC0+nC1+nC2+⋯+nCn
4. Final Calculation:
According to the property of binomial coefficients, the sum of all coefficients in the expansion of (1+x)n (where x=1) is:
∑r=0nnCr=2n
Correct Option: (b) 2n