JEE Main 2026 (Online) 28th January Evening ShiftQuestion 1
$\frac{6}{3^{26}}+\frac{10\cdot1}{3^{25}}+\frac{10\cdot2}{3^{24}}+\frac{10\cdot2^2}{3^{23}}+\ldots+\frac{10\cdot2^{24}}{3}$ is equal to :
Sequence and Series · Geometric Progression
Solution
Formulas used. (1) Sum of a G.P.: $\sum_{j=0}^{n-1}x^j=\frac{x^n-1}{x-1}$(2) $a^m b^m=(ab)^m$Step 1 - Clear the denominators. Multiply the whole expression by $3^{26}$. The first term $\dfrac6{3^{26}}$ becomes $6$. The term $\dfrac{10\cdot2^{j}}{3^{25-j}}$ becomes $10\cdot2^j\cdot3^{\,j+1}$, for $j=0,1,\dots,24$. $$3^{26}E=6+\sum_{j=0}^{24}10\cdot2^j\cdot3^{j+1}.$$ Step 2 - The G.P. $$\sum_{j=0}^{24}10\cdot2^j3^{j+1}=30\sum_{j=0}^{24}6^j=30\cdot\frac{6^{25}-1}{6-1}=6\left(6^{25}-1\right)=6^{26}-6 .$$ Step 3 - Combine. $$3^{26}E=6+6^{26}-6=6^{26}\ \Longrightarrow\ E=\frac{6^{26}}{3^{26}}=2^{26}.$$ Answer: (D)