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JEE Main 202224 JuneEvening Shift
Exam
JEE Main
Year
2022
Date
24 June
Shift
Evening Shift
Subject
Mathematics
Chapter
Binomial Theorem

JEE Main 2022 · 24 June · Evening Shift · Mathematics · Binomial Theorem

The remainder on dividing 1 + 3 + 3 ^ 2 + 3 ^ 3 + + 3 ^ 2021 by 50...

The remainder on dividing 1+3+32+33+…+320211 + 3 + {3}^{2} + {3}^{3} + \ldots + {3}^{2021} by 5050 is _____.

Correct answer

0

Step-by-step solution

Let S=1+3+32+⋯⋅32021S = 1 + 3 + {3}^{2} + ⋯ \cdot {3}^{2021}

=32022−13−1=32022−12= \dfrac{{3}^{2022} - 1}{3 - 1} = \dfrac{{3}^{2022} - 1}{2}

Simplifying by using binomial expansion we get(32)1011−12=(10−1)1011−12\dfrac{{\left({3}^{2}\right)}^{1011} - 1}{2} = \dfrac{( 10 - 1{)}^{1011} - 1}{2}

⇒1011C0101011−1011C1101010+…+C10101011101−C10111011×1−12⇒ \dfrac{{}^{1011}{C}_{0}{10}^{1011}{-}^{1011}{C}_{1}{10}^{1010} + \ldots + C1010 1011 {10}^{1} - C1011 1011 \times 1 - 1}{2}

100q+1011×10−22\dfrac{100q + 1011 \times 10 - 2}{2}

⇒100q+10110−22=100q+101082=50q+5054⇒ \dfrac{100q + 10110 - 2}{2} = \dfrac{100q + 10108}{2} = 50q + 5054

Now dividing 50q+505450q + 5054 by 5050

We get 44 as remainder.

Hence option B is correct.

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