Find the value of n, so that an+1+bn+1an+bn is the geometric mean betw

1.	Find the value of n, so that an+1+bn+1an+bn is the geometric mean between a and b Or If f is a function satisfying f(x+y)=f(x)f(y) for all x,y∈N such that f(1)=3 and ∑nx=1f(x)=120 find the value of n.
Answer» Find the value of n, so that $\frac{a^{n + 1} + b^{n + 1}}{a^{n} + b^{n}}$ is the geometric mean between a and b Or If f is a function satisfying $f (x + y) = f (x) f (y)$ for all $x, y \in N$ such that f(1)=3 and $\sum_{x = 1}^{n} f (x) = 120$ find the value of n.

Find the value of n, so that an+1+bn+1an+bn is the geometric mean between a and b Or If f is a function satisfying f(x+y)=f(x)f(y) for all x,y∈N such that f(1)=3 and ∑nx=1f(x)=120 find the value of n.

Answer»

Find the value of n, so that $\frac{a^{n + 1} + b^{n + 1}}{a^{n} + b^{n}}$ is the geometric mean between a and b

If f is a function satisfying $f (x + y) = f (x) f (y)$ for all $x, y \in N$ such that f(1)=3 and $\sum_{x = 1}^{n} f (x) = 120$ find the value of n.