JEE Mains · Maths · STD 11 - 7. binomial theoram
The remainder on dividing \(1+3+3^{2}+3^{3}+\ldots+3^{2021}\) by \(50\) is
- A \(5\)
- B \(4\)
- C \(2\)
- D \(6\)
Answer & Solution
Correct Answer
(B) \(4\)
Step-by-step Solution
Detailed explanation
\(\frac{1 .\left(3^{2022}-1\right)}{2}=\frac{9^{1011}-1}{2}\) \(=\frac{(10-1)^{1011}-1}{2}\) \(=\frac{100 \lambda+10110-1-1}{2}\) \(=50 \lambda+\frac{10108}{2}\) \(=50 \lambda+5054\) \(=50 \lambda+50 \times 101+4\) \(\operatorname{Rem}(50)=4 .\)
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