JEE Main 2022 — Binomial Theorem Question with Solution

Given,

$$1+3+3^2+3^3+.....+3^{2021}$$

$$=3^0+3^1+3^2+3^3+....+3^{2021}$$

This is a G.P with common ratio = 3

$$\therefore$$ Sum $$=\frac{1(3^{2022}-1)}{3-1}$$

$$=\frac{3^{2022}-1}{2}$$

$$=\frac{{(3^2)}^{2011}-1}{2}$$

$$=\frac{{(10-1)}^{1011}-1}{2}$$

$$=\frac{\left[{}^{1011}{C}_0.{10}^{1011}-{}^{1011}{C}_1.{10}^{1010}+.....-{}^{1011}{C}_{1009}.{(10)}^2+{}^{1011}{C}_{1010}.10-{}^{1011}{C}_{1011}\right]-1}{2}$$

$$=\frac{{10}^2\left[{}^{1011}{C}_0.{(10)}^{1009}-{}^{1011}{C}_1.(1008)+.....{}^{1011}{C}_{1009}\right]+10110-1-1}{2}$$

$$=\frac{100k+10110-2}{2}$$

$$=\frac{100k+10108}{2}$$

$$=50k+5054$$

$$=50k+50\times 101+4$$

$$=50[k+101]+4$$

$$=50k^{\prime }+4$$

$$\therefore$$ By dividing 50 we get remainder as 4.