Mathematics · 2020
JEE Main · 3 September 2020, Shift 1 · Q53
The value of (2 ·^1 P_0-3 ·^2 P_1+4 ·^3 P_2-…. up to 51^th term )+(1!-2!+3!-…. up to 51^th term) is equal to:
The value of $\displaystyle \left(2 \cdot{ }^1 \mathrm{P}_0-3 \cdot{ }^2 \mathrm{P}_1+4 \cdot{ }^3 \mathrm{P}_2-\ldots\right.$ up to $\displaystyle 51^{\text {th }}$ term $\displaystyle )+\left(1!-2!+3!-\ldots\right.$ up to $\displaystyle 51^{\text {th }}$ term) is equal to :
Official answer
(1)
$\displaystyle 1+(52)!$
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JEE Main 2020 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.