If-f-x-x-n-then-the-value-of-f-1-f-1-1-1-f-2-1-2-f-n-1-n-where-f-r-x-denotes-the-rth-order-derivative-of-f-x-with-respect-to-x-is-

Question Number 8155 by 314159 last updated on 02/Oct/16

If f (x) = x^{n}, then the value of

f (1) + \frac{f^{1} (1)}{1} + \frac{f^{2} (1)}{2!} + \dots + \frac{f^{n} (1)}{n!}, where

f^{r} (x) denotes the r th order derivative of

f (x) with respect to x, is

Commented by sou1618 last updated on 02/Oct/16

f^{1} (x) = {n x}^{n - 1}

f^{2} (x) = n (n - 1) x^{n - 2}

\dots

f^{r} (x) = n (n - 1) (n - 2) \dots (n - r + 1) x^{n - r}

/ / / / / / / / / /

f i n d S = 1 + \frac{n}{1!} + \frac{n (n - 1)}{2!} + \frac{n (n - 1) (n - 2)}{3!} + \dots + \frac{n!}{n!}

\Leftrightarrow

S = 1 + \frac{n!}{1! (n - 1)!} + \frac{n!}{2! (n - 2)!} + \frac{n!}{3! (n - 3)!} + \dots + \frac{n!}{n! (n - n)!}

=_{n} C_{0} +_{n} C_{1} +_{n} C_{2} +_{n} C_{3} + \dots +_{n} C_{n}

*_{n} C_{r} = (\begin{matrix} n \\ r \end{matrix}) = \frac{n!}{r! (n - r)!}

= \underset{k = 0}{\sum^{n}}_{n} C_{k}

/ / / / / / / / / / / /

s e t g (x) = {(1 + x)}^{n}

g (x) = \underset{k = 0}{\sum^{n}} (1^{k} x^{n - k}_{n} C_{k}) .

g (1) = \underset{k = 0}{\sum^{n}}_{n} C_{k} = S

g (1) = {(1 + 1)}^{n} = 2^{n}

s o

S = 2^{n}

Commented by 314159 last updated on 02/Oct/16

T h a n k s!

Answered by prakash jain last updated on 02/Oct/16

Answer in comments