If fx-xn-4- then f1-f-1-1-f-1-2-f-1-3-fn1-n-

The correct option is B 2^n+4 f(x)=1+4 f'(x)=n(x)^n 1 #160; f'(1)=n f”(x)=n(x)^n 1x^n 2 #160; f”(1) = n (n 1) f”'(x)=n(n ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Monotonically Increasing Functions

If fx=xn+4,...

Question

If $f (x) = x^{n} + 4$ , then $f (1) + \frac{f^{^{'}} (1)}{1!} + \frac{f^{^{''}} (1)}{2!} + \frac{f^{^{'''}} (1)}{3!} + \dots + \frac{f^{n} (1)}{n!} =$

2^{n - 1}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

2^{n} + 4

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

2^{n}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

1 + \frac{1}{1!} + \frac{1}{2!} + \dots + \frac{1}{n!}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

Suggest Corrections

Similar questions

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

f (0) + f^{'} (0) + \frac{1}{2!} f^{''} (0) + \dots + \frac{1}{n!}

f^{n} (0)

2^{n}

f (x) = x^{n} + 4

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

2^{n} + 4

f (x) = x^{n}

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

{(1 + x)}^{n} = n \sum r = 0^{n} C_{r} x^{r}

n \sum r = 0^{n} C_{r} = 2^{n}

f (x) = x^{n}

f^{'} (x) = r!^{n} C_{r} x^{n - r}

r^{t h}

f (x)

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

{(1 + x)}^{n}

2^{n}

\frac{1^{4}}{1.3} + \frac{2^{4}}{3.5} + \frac{3^{4}}{5.7} + . . . + \frac{n^{4}}{(2 n - 1) . (2 n + 1)} = \frac{1}{48} f (n) + \frac{n}{16 (2 n + 1)}

f (n)

f (n) = \frac{1}{n} {(2 n + 1) (2 n + 2) \dots (2 n + n)}^{1 / n}

lim n \to \infty f (n)

Join BYJU'S Learning Program