If $f (x) = x^{n}$ , then the value of
$f (1) - \frac{f ^{'} ( 1 )}{1 !} + \frac{f ^{''} ( 1 )}{2 !} - \frac{f ^{'''} ( 1 )}{3 !} + . . . + \frac{( - 1 ) ^{n} f ^{n} ( 1 )}{n !}$ is

Question

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Answer 1

$f (x) = x^{n} \Rightarrow f (1) = 1$
$f^{'} (x) = n x^{n - 1} \Rightarrow f^{'} (1) = n$
$f^{''} (x) = n (n - 1) x^{n - 2} \Rightarrow f^{''} (1) = n (n - 1)$
... ... ... ...
... ... ... ...
$f^{n} (x) = n (n - 1) (n - 2) . . . 2.1$
$f^{n} (1) = n (n - 1) (n - 2) . . . 2.1$
$f (1) - \frac{f ^{'} ( 1 )}{1 !} + \frac{f ^{''} ( 1 )}{2 !} - \frac{f ^{'''} ( 1 )}{3 !} + . . . + \frac{( - 1 ) ^{n} f ^{n} ( 1 )}{n !}$
$= 1 - \frac{n}{1 !} + \frac{n ( n - 1 )}{2 !} - \frac{n ( n - 1 ) ( n - 2 )}{3 !} + . . . + \frac{( - 1 ) ^{n} n ( n - 1 ) ( n - 2 ) . . . 2 . 1}{n !}$
$= (1 - 1)^{n}$ ........... [Using binomial expansion as $(1 - x)^{n} =^{n} C_{0} 1^{n} (- x)^{0} +^{n} C_{1} 1^{n - 1} (- x)^{1} + . . . . . +^{n} C_{n} (- x)^{n}]$
$= 0$

Answer 2

f (x) = x^{n} \Rightarrow f (1) = 1

f^{'} (x) = n x^{n - 1} \Rightarrow f^{'} (1) = n

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

... ... ... ...

f^{n} (x) = n (n - 1) (n - 2) . . . 2.1

f^{n} (1) = n (n - 1) (n - 2) . . . 2.1

f (1) - \frac{f ^{'} ( 1 )}{1 !} + \frac{f ^{''} ( 1 )}{2 !} - \frac{f ^{'''} ( 1 )}{3 !} + . . . + \frac{( - 1 ) ^{n} f ^{n} ( 1 )}{n !}

= 1 - \frac{n}{1 !} + \frac{n ( n - 1 )}{2 !} - \frac{n ( n - 1 ) ( n - 2 )}{3 !} + . . . + \frac{( - 1 ) ^{n} n ( n - 1 ) ( n - 2 ) . . . 2 . 1}{n !}

= (1 - 1)^{n}

........... [Using binomial expansion as

(1 - x)^{n} =^{n} C_{0} 1^{n} (- x)^{0} +^{n} C_{1} 1^{n - 1} (- x)^{1} + . . . . . +^{n} C_{n} (- x)^{n}]

= 0