Question

Let $f\left(1\right)=1andf\left(n\right)=2\sum _{r=1}^{n-1}f\left(x\right)$, then $\sum _{n=1}^{m}f\left(n\right)$ is equal to

• A. $3^m-1$
• B. $3^m$
• C. $3^{m-1}$
• D. None of these

Answer 1

The correct option is C $3^{m-1}$

$f\left(1\right)=1f\left(2\right)=2\stackrel{1}{\sum _{x=1}}f\left(x\right)=2f\left(1\right)=2\times 1=2f\left(3\right)=2\stackrel{2}{\sum _{x=1}}f\left(x\right)=2f\left(1\right)+2f\left(2\right)=2\times 1+2\times 2=6f\left(4\right)=2\stackrel{3}{\sum _{x=1}}f\left(x\right)=2\left[f\right(1)+f(2)+f(3\left)\right]=2[1+2+6]=18$

The sequence is $=1+2+6+18+......$

$\stackrel{m}{\sum _{n=1}}f\left(n\right)$$=1+2(1+3+9+......)=1+2\frac{\left(1\right)(3^{m-1}-1)}{2}$

$\stackrel{m}{\sum _{n=1}}f\left(n\right)=3^{m-1}$