The remainder on dividing 1+3+3^{2}+3^{3}+ldo | vedclass

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(B) The given sum is a geometric progression with $a=1$,$r=3$,and $n=2022$ terms.Sum $S = \frac{1(3^{2022}-1)}{3-1} = \frac{3^{2022}-1}{2}$.We can wri

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(B) The given sum is a geometric progression with $a=1$,$r=3$,and $n=2022$ terms.Sum $S = \frac{1(3^{2022}-1)}{3-1} = \frac{3^{2022}-1}{2}$.We can write $3^{2022} = (3^2)^{1011} = 9^{1011} = (10-1)^{1011}$.Using the binomial expansion,$(10-1)^{1011} = \binom{1011}{0} 10^{1011} - \binom{1011}{1} 10^{1010} + \ldots + \binom{1011}{1010} 10^1 (-1)^{1010} + (-1)^{1011}$.$(10-1)^{1011} = 100k + 1011(10)(-1) - 1 = 100k - 10110 - 1 = 100k - 10111$.Thus,$S = \frac{100k - 10111 - 1}{2} = \frac{100k - 10112}{2} = 50k - 5056$.$S = 50k - 5050 - 6 = 50(k-101) - 6$.Since the remainder must be positive,we write $S = 50(k-102) + 50 - 6 = 50(k-102) + 4$.Therefore,the remainder is $4$.

The remainder on dividing $1+3+3^{2}+3^{3}+\ldots+3^{2021}$ by $50$ is

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