Method of DifferentiationhardFree

Method of Differentiation — JEE Maths practice question

JEE Maths question with a full step-by-step solution.

Question
If f(x)=(2015+x)nf(x)=(2015+x)^{n} where xx is a real variable and nn is a positive integer, then the value of f(0)+f′(0)1!+f′′(0)2!+⋯+f(n−1)(0)(n−1)!f(0)+\dfrac{f'(0)}{1!}+\dfrac{f''(0)}{2!}+\cdots+\dfrac{f^{(n-1)}(0)}{(n-1)!} is
A(2015)n(2015)^{n}
B(2015)n−1(2015)^{n}-1
C(2016)n(2016)^{n}
D(2016)n−1(2016)^{n}-1correct
Solution
Step 1: Compute the kk-th derivative of f(x)=(2015+x)nf(x)=(2015+x)^{n}:
f(k)(x)=n!(n−k)!(2015+x)n−kf^{(k)}(x)=\dfrac{n!}{(n-k)!}(2015+x)^{n-k}
 Step 2: Evaluate at x=0x=0 and divide by k!k!:
f(k)(0)k!=n!(n−k)!⋅k!⋅2015n−k=(nk)⋅2015n−k\dfrac{f^{(k)}(0)}{k!}=\dfrac{n!}{(n-k)!\cdot k!}\cdot 2015^{n-k}=\binom{n}{k}\cdot 2015^{n-k}
 Step 3: The required sum is:
S=∑k=0n−1f(k)(0)k!=∑k=0n−1(nk)⋅2015n−k⋅1kS=\sum_{k=0}^{n-1}\dfrac{f^{(k)}(0)}{k!}=\sum_{k=0}^{n-1}\binom{n}{k}\cdot 2015^{n-k}\cdot 1^{k}
 Step 4: By the Binomial theorem, the complete sum up to k=nk=n equals:
∑k=0n(nk)⋅2015n−k⋅1k=(2015+1)n=2016n\sum_{k=0}^{n}\binom{n}{k}\cdot 2015^{n-k}\cdot 1^{k}=(2015+1)^{n}=2016^{n}
 Step 5: The required sum SS excludes only the last term at k=nk=n:
(nn)⋅20150⋅1n=1\binom{n}{n}\cdot 2015^{0}\cdot 1^{n}=1
 Step 6: Therefore:
S=2016n−1S=2016^{n}-1
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