Question

If $\underset{x\to 2}{lim}\frac{\sum _{r=1}^n{x}^r-\sum _{r=1}^n{2}^r}{x-2}=f\left(n\right),$ then which of the following is/are CORRECT?

• A. $f\left(2020\right)=1999(2{)}^{2020}+1$
• B. $f\left(2020\right)=505(2{)}^{2022}+1$
• C. $f\left(6\right)=321$
• D. $f\left(10\right)=9217$

Answer 1

Answer: (A), (C), (D)

$f\left(10\right)=9217$

Given : $\underset{x\to 2}{lim}\frac{\sum _{r=1}^n{x}^r-\sum _{r=1}^n{2}^r}{x-2}$

Now,
$f\left(n\right)=\underset{x\to 2}{lim}\frac{(x-2)}{x-2}+\frac{(x^2-2^2)}{x-2}+\cdots +\frac{(x^n-2^n)}{x-2}\Rightarrow f\left(n\right)=1+2\cdot 2+3\cdot 2^2+\cdots +n\cdot 2^{n-1}\cdots \left(i\right)\Rightarrow 2f\left(n\right)=2+2\cdot 2^2+\cdots +(n-1)\cdot 2^n+n\cdot 2^n\cdots \left(ii\right)$
Subtracting $\left(i\right)$ and $\left(ii\right)$
$\Rightarrow -f\left(n\right)=(1+2+2^2+\cdots 2^{n-1})-n\cdot 2^n\Rightarrow -f\left(n\right)=\frac{1(2^n-1)}{2-1}-n{2}^n\Rightarrow -f\left(n\right)=2^n-1-n{2}^n\therefore f\left(n\right)=(n-1)2^n+1$

Hence,
$f\left(2020\right)=2019(2{)}^{2020}+1f(6)=5\cdot 2^6+1=321f(10)=9\cdot 2^{10}+1=9217$