Question

Let $f\left(1\right)=1$ and $f\left(n\right)=2{\sum }_{r=1}^{n-1}f\left(r\right)$. Then ${\sum }_{n=1}^{m}f\left(n\right)$. is equal to

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

Answer 1

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

$\begin{array}{c}f\left(1\right)=1\\ f\left(2\right)=2{\sum }_{x=1}^{1}f\left(x\right)=2f\left(1\right)=2\\ f\left(3\right)=2{\sum }_{x=1}^{2}f\left(x\right)=2\left(f\right(1)+f(2\left)\right)=6\\ \text{The sequence is}1+2+6+18+\dots \\ \begin{array}{cc}\sum _{n-1}f\left(n\right)& =1+2+6+18t\\ & =1+2(1+3+9+\dots )\\ & =1+2\left(\frac{3^{m-1}-1}{3-1}\right)\\ & =1+3^{m-1}-1\\ & =3^{m-1}\end{array}\end{array}$

option C is correct.