Question

$\sum _{k=1}^{\infty }\frac{6^k}{(3^{2k+1}+2^{2k+1})-(3^k{2}^{k+1}+2^k{3}^{k+1})}$ is equal to

• A. $3$
• B. $\frac{1}{3}$
• C. $\frac{4}{3}$
• D. $\frac{6}{5}$

Answer 1

Answer: (A)

$3$

${\sum }_{k=1}^{\infty }\frac{6^k}{(3^{2k+1}+2^{2k+1})-(3^k{2}^{k+1}+2^k{3}^{k+1})}={\sum }_{k=1}^{\infty }\frac{6^k}{3\cdot 3^{2k}-3\cdot 3^k{2}^k+2^2{2}^{2k}-2\cdot 3^k{2}^k}={\sum }_{k=1}^{\infty }\frac{6^k}{3\cdot 3^k(3^k-2^k)-2\cdot 2^k(3^k-2^k)}={\sum }_{k=1}^{\infty }\frac{6^k}{(3^k-2^k)(3^{k+1}-2^{k+1})}\frac{6^k}{(3^k-2^k)(3^{k+1}-2^{k+1})}=\frac{3^{k}A}{(3^k-2^k)}+\frac{3^{k}B}{(3^{k+1}-2^{k+1})}\frac{6^k}{(3^k-2^k)(3^{k+1}-2^{k+1})}=\frac{(3A+B)3^{2k}B-(2A+B)6^k}{(3^k-2^k)(3^{k+1}-2^{k+1})}3A+B=0\Rightarrow 3A=-B\left(1\right)2A+B=-12A-3A=-1from\left(1\right)-A=-1A=1\Rightarrow B=-3A=-3{\sum }_{k=1}^{\infty }\frac{6^k}{(3^k-2^k)(3^{k+1}-2^{k+1})}=\sum \frac{3^k}{(3^k-2^k)}-\sum \frac{3^{k+1}}{(3^{k+1}-2^{k+1})}{T}_1=\frac{3}{3-2}-\frac{3^2}{3^2-2^2}{T}_2=\frac{3^2}{3^2-2^2}-\frac{3^3}{3^3-2^3}{T}_n=\frac{3^k}{(3^k-2^k)}-\frac{3^{k+1}}{(3^{k+1}-2^{k+1})}{S}_n=3-\frac{3^{n+1}}{3^{n+1}-2^{n+1}}\underset{n\to \infty }{lim}{S}_n=3$