If f(x) = 1 + nx + (n(n-1)/2)x2+(n(n-1)(n-2)/6)x3+.........+xn then f''(1) =

Question

If f(x) = 1 + nx + (n(n-1)/2)x2+(n(n-1)(n-2)/6)x3+.........+xn then f''(1) = (A) n(n-1)2n (B) n(n-1)2n (C) (n-1)2n-1 (D) ⇒ f prime prime(1)=n(n-1) 2n-2

Tardigrade Admin · Accepted Answer

Given, f(x)= 1+n x+(n(n-1)/2 !) x2 +(n(n-1)(n-2)/3 !) x3+ ldots+xn ⇒ f(x)=(1+x)n ⇒ f'(x)=n(1+x)n-1 ⇒ f''(x)=n(n-1)(1+x)n-2 ⇒ f''(1)=n(n-1) 2n-2

Q. If $f (x) = 1 + n x + \frac{n ( n - 1 )}{2} x^{2} + \frac{n ( n - 1 ) ( n - 2 )}{6} x^{3} + ......... + x^{n}$ then $f^{''} (1)$ =

$n (n - 1)^{2 n}$

$n (n - 1) 2^{n}$

$(n - 1) 2^{n - 1}$

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

Given, $f (x) = 1 + n x + \frac{n ( n - 1 )}{2 !} x^{2} + \frac{n ( n - 1 ) ( n - 2 )}{3 !} x^{3} + \dots + x^{n}$
$\Rightarrow f (x) = (1 + x)^{n}$
$\Rightarrow f^{'} (x) = n (1 + x)^{n - 1}$
$\Rightarrow f^{''} (x) = n (n - 1) (1 + x)^{n - 2}$
$\Rightarrow f^{''} (1) = n (n - 1) 2^{n - 2}$

Q. If f(x)=1+nx+2n(n−1)​x2+6n(n−1)(n−2)​x3+.........+xn then f′′(1) =

n(n−1)2n