QJEEM23Mae3b3: JEE Main 2023 Mathematics - Class 11 JEEM23Mae3b3 Solution: If , then is equal to ___________.

Question

Verified Solution • 2025–26 Session moderate

If

$(20)^{19}+2(21)(20)^{18}+3(21)^{2}(20)^{17}+\ldots+20(21)^{19}=k(20)^{19}$ ,

then $k$ is equal to ___________.

Accepted Answer

\begin{aligned} &amp;(20)^{19}+2(21)(20)^{18}+3(21)^2(20)^{17} \\ &amp; \quad+\ldots \ldots+20(21)^{19}=k(20)^{19} \\\\ &amp; \Rightarrow(20)^{19}\left[1+2\left(\frac{21}{20}\right)+3\left(\frac{21}{20}\right)^2+\ldots+20\left(\frac{21}{20}\right)^{19}\right]=k(20)^{19} \\\\ &amp; \Rightarrow k=1+2\left(\frac{21}{20}\right)+3\left(\frac{21}{20}\right)^2+\ldots+20\left(\frac{21}{20}\right)^{19} ..........(i)\end{aligned}

Now,

\begin{aligned} k\left(\frac{21}{20}\right)=\left(\frac{21}{20}\right) &amp; +2\left(\frac{21}{20}\right)^2+3\left(\frac{21}{20}\right)^3 +\ldots+20\left(\frac{21}{20}\right)^{20}\end{aligned}

........(ii)

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

\begin{aligned}&amp; k\left(\frac{-1}{20}\right)=1+\frac{21}{20}+\left(\frac{21}{20}\right)^2+\ldots \ldots+\left(\frac{21}{20}\right)^{19}-20\left(\frac{21}{20}\right)^{20} \\\\&amp; \Rightarrow k\left(\frac{-1}{20}\right)=\frac{1\left(\left(\frac{21}{20}\right)^{20}-1\right)}{\frac{21}{20}-1}-20\left(\frac{21}{20}\right)^{20} \\\\&amp; \Rightarrow k\left(\frac{-1}{20}\right)=20\left(\left(\frac{21}{20}\right)^{20}-1\right)-20\left(\frac{21}{20}\right)^{20} \\\\&amp; =20\left(\frac{21}{20}\right)^{20}-20-20\left(\frac{21}{20}\right)^{20} \\\\&amp; \Rightarrow k\left(\frac{-1}{20}\right)=-20 \\\\&amp; \Rightarrow k=400\end{aligned}

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