If [(20)^19+2(21)(20)^18+3(21)^2(20)^17++20(21)^19=k(20)^19], then k is equal to _________…

Question

If 
$$(20)^{19}+2(21)(20)^{18}+3(21)^{2}(20)^{17}+\ldots+20(21)^{19}=k(20)^{19}$$, 
then $k$ is equal to \_\_\_\_\_\_\_\_\_\_\_.

Answer: 400

Accepted Answer

$\begin{gathered} (20{)}^{19}+2(21)(20{)}^{18}+3(21{)}^{2}(20{)}^{17} \ \, +\ldots \ldots +20(21{)}^{19}=k(20{)}^{19} \ \Rightarrow (20{)}^{19}[1+2(\frac{21}{20})+3{(\frac{21}{20})}^{2}+\ldots +20{(\frac{21}{20})}^{19}]=k(20{)}^{19} \ \Rightarrow k=1+2(\frac{21}{20})+3{(\frac{21}{20})}^{2}+\ldots +20{(\frac{21}{20})}^{19}..........(i) \end{gathered}$

Now,

$\begin{gathered} k(\frac{21}{20})=(\frac{21}{20}) +2{(\frac{21}{20})}^{2}+3{(\frac{21}{20})}^{3}+\ldots +20{(\frac{21}{20})}^{20} \end{gathered}$ ........(ii)

On subtracting Equation (ii) from Equation (i), we get

$\begin{gathered} k(\frac{-1}{20})=1+\frac{21}{20}+{(\frac{21}{20})}^{2}+\ldots \ldots +{(\frac{21}{20})}^{19}-20{(\frac{21}{20})}^{20} \ \Rightarrow k(\frac{-1}{20})=\frac{1({(\frac{21}{20})}^{20}-1)}{\frac{21}{20}-1}-20{(\frac{21}{20})}^{20} \ \Rightarrow k(\frac{-1}{20})=20({(\frac{21}{20})}^{20}-1)-20{(\frac{21}{20})}^{20} \ =20{(\frac{21}{20})}^{20}-20-20{(\frac{21}{20})}^{20} \ \Rightarrow k(\frac{-1}{20})=-20 \ \Rightarrow k=400 \end{gathered}$

Solutions · 1