Question

Assertion :Let $f\left(x\right)=x^n$ & $f^{\prime }\left(x\right)=r{!}^n{C}_r{x}^{n-r}$ denotes the $r^{th}$ order derivative of $f\left(x\right)$ then $f\left(1\right)+\frac{f^{\prime }\left(1\right)}{1!}+\frac{f^{\prime \prime }\left(1\right)}{2!}+.....+\frac{f^n\left(1\right)}{n!}=2^n$ Reason: The sum of binomial coefficients in the expansion of ${(1+x)}^n$ is $2^n$

• A. Both Assertion & Reason are individually true & Reason is correct explanation of Assertion,
• B. Both Assertion & Reason are individually true but Reason is not the ,correct (proper) explanation of Assertion,
• C. Assertion is true but Reason is false,
• D. Assertion is false but Reason is true.

Answer 1

The correct option is B Both Assertion & Reason are individually true but Reason is not the ,correct (proper) explanation of Assertion,

$f\left(x\right)+\frac{f^{\prime }\left(x\right)}{1!}+\frac{f^{\prime \prime }\left(x\right)}{2!}+....+\frac{f^n\left(x\right)}{n!}$
$=x^n+\frac{n{x}^{n-1}}{1!}+\frac{n(n-1)x^{n-2}}{2!}+...+\frac{n!x}{(n-1)!}+\frac{n!}{n!}$
$={}^n{C}_0{x}^n+{}^n{C}_1{x}^{n-1}+{}^n{C}_2{x}^{n-2}+...+{}^n{C}_n{x}^0$
$=(x+1{)}^n$
Now substituting $x=1$, we get
The sum of the series as $2^n$
For
$(1+x{)}^n$ sum of the coefficients is $2^n$ however this is not the proper explanation of the assertion.