Method of DifferentiationhardFree

Method of Differentiation: Real Variable Positive Integer Value

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

Question

f(x)=(2015+x)^{n}

where

x

is a real variable and

n

is a positive integer, then the value of

f(0)+\dfrac{f'(0)}{1!}+\dfrac{f''(0)}{2!}+\cdots+\dfrac{f^{(n-1)}(0)}{(n-1)!}

(2015)^{n}

(2015)^{n}-1

(2016)^{n}

(2016)^{n}-1

correct

Solution

Step 1: Compute the

k

-th derivative of

f(x)=(2015+x)^{n}

f^{(k)}(x)=\dfrac{n!}{(n-k)!}(2015+x)^{n-k}

Step 2: Evaluate at

x=0

and divide by

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=\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=n

equals:

\sum_{k=0}^{n}\binom{n}{k}\cdot 2015^{n-k}\cdot 1^{k}=(2015+1)^{n}=2016^{n}

Step 5: The required sum

S

excludes only the last term at

k=n

\binom{n}{n}\cdot 2015^{0}\cdot 1^{n}=1

Step 6: Therefore:

S=2016^{n}-1

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