Question

If the arithmetic mean between $a$ and $b$ is $\frac{a^{n+1}+b^{n+1}}{a^n+b^n}$, then $n$, is equal to

• A. $0$
• B. $1$
• C. $-1$
• D. $\frac{1}{2}$

Answer 1

Answer: (A)

$0$

$2\left[\right(a{)}^{n+1}+\left(b{)}^{n+1}\right]=(a+b)\left(\right(a{)}^n+\left(b{)}^n\right)$
$2(a{)}^{n+1}+2(b{)}^{n+1}=(a{)}^{n+1}+a(b{)}^n+(a{)}^{n}b+(b{)}^{n+1}$
$(a{)}^{n+1}+(b{)}^{n+1}-a(b{)}^n-(a{)}^{n}b=0$
$\left(\right(a{)}^n-\left(b{)}^n\right)(a-b)=0$
$(a{)}^n=(b{)}^n$ (This is only possible when $n=0$)
Hence, option A.