First slide

Introduction to real valued functions

Question

f is a function defined as $\sum_{k = 1}^{n} f (a + k) = 16 (2^{n} - 1)$ and $f (x + y) = f (x) . f (y)$ and f(1) = 2 then integral value of a

Moderate

A

3
B

0
C

2
D

1

Solution

$f (x + y) = f (x) f (y) x = y = 1 \Rightarrow f (2) = f (1) f (1) = 4 x = 1, y = 2 \Rightarrow f (3) = f (1) f (2) = 8 \Rightarrow f (x) = 2^{x} \sum_{k = 1}^{n} 2^{a + k} = 16 (2^{n} - 1) 2^{a} (2 + 2^{2} + 2^{3} + . . . . . . . . . . . + 2^{n}) = 16 (2^{n} - 1) u s e G . P f o r m u l a$

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

Doubts App

OR

SCAN ME

Doubts App

Doubts App