Question

Let a cricket player played $n(n>1)$ matches during his career. If $T_r$ represents the runs made by the player in $r^{\text{th}}$ match such that $T_1=6$ and $T_r=3T_{r–1}+6^r$ for $2\le r\le n$, then the runs scored by him in $100$ matches is

• A. $\frac{3}{5}[4\cdot 6^{100}-5\cdot 3^{100}-1]$
• B. $\frac{2}{5}[4\cdot 6^{100}-5\cdot 3^{100}+1]$
• C. $\frac{3}{5}[4\cdot 6^{99}-5\cdot 3^{99}+1]$
• D. $\frac{3}{5}[4\cdot 6^{100}-5\cdot 3^{100}+1]$

Answer 1

The correct option is D $\frac{3}{5}[4\cdot 6^{100}-5\cdot 3^{100}+1]$

Given : $T_r=3T_{r–1}+6^r$
$\Rightarrow T_r-3T_{r–1}=6^r\Rightarrow \frac{T_r}{3^r}-\frac{T_{r-1}}{3^{r-1}}=2^r\Rightarrow \sum _{r=2}^n(\frac{T_r}{3^r}-\frac{T_{r-1}}{3^{r-1}})=\frac{T_2}{3^2}-\frac{T_1}{3}+\frac{T_3}{3^3}-\frac{T_2}{3^2}+\frac{T_4}{3^4}-\frac{T_3}{3^3}⋮+\frac{T_n}{3^n}-\frac{T_{n-1}}{3^{n-1}}\Rightarrow \sum _{r=2}^n(\frac{T_r}{3^r}-\frac{T_{r-1}}{3^{r-1}})=\frac{T_n}{3^n}-\frac{T_1}{3}$
Therefore, $\frac{T_n}{3^n}-\frac{T_1}{3}=\sum _{r=2}^n{2}^r$
$\Rightarrow \frac{T_n}{3^n}-2=2^2(2^{n-1}-1)\Rightarrow \frac{T_n}{3^n}=2(2^n-1)\Rightarrow T_n=2(6^n-3^n)$

So, $S_n=2(\sum _1^n{6}^r-\sum _1^n{3}^r)$
$\Rightarrow S_n=2\left[\frac{6}{5}\right(6^n-1)-\frac{3}{2}(3^n-1\left)\right]\Rightarrow S_n=\frac{3}{5}[4\cdot 6^n-5\cdot 3^n+1]$
Hence, the runs scored by the player
$=\frac{3}{5}[4\cdot 6^{100}-5\cdot 3^{100}+1]$