Question

Let f be a function such that $f\left(3\right)=1$ and $f\left(3x\right)=x+f(3x-3)$ for all $x$. Then find the value of $f\left(300\right)$.

• A. $5500$
• B. $10100$
• C. $5050$
• D. $1,71,700$

Answer 1

Answer: (D)

$1,71,700$

$f\left(3\right)=1$

$f\left(3x\right)=x+f(3x-3)$

Substitute $x=1$

$f\left(3\right)=1+f\left(0\right)$

$f\left(0\right)=0$

put $x=2$

$f\left(6\right)=2+f\left(3\right)=3$

put $x=3$

$f\left(9\right)=3+f\left(6\right)=3+3=6$

put $x=4$

$f\left(12\right)=4+6=10$

Hence $f\left(300\right)=1+3+6+10+{.100}^{th}$ term

$S=1+3+6+10....+T_n$

$S=1+(1+2)+(1+2+3)+10.......+T_n$

$T_r=1+2+3+4.....rterms=\frac{r(r+1)}{2}$

$s=\sum t_r=\sum \frac{r^2+r}{2}$

$=\frac{1}{2}{(\frac{n(n+1)(2n+1)}{6}+\frac{n(n+1)}{2})}_{n=100}$

$\therefore f\left(300\right)=\frac{1}{2}(\frac{\mathrm{100.101.201}}{6}+\frac{100.101}{2})=1,71,700$