Mathematics · 2021
JEE Main · 18 March 2021, Shift 2 · Q84
If Σ_r=1^10 r!(r^3+6 r^2+2 r+5)=α(11!), then the value of α is equal to \ \ \ \.
If $\displaystyle \sum_{r=1}^{10} r!\left(r^3+6 r^2+2 r+5\right)=\alpha(11!)$,
then the value of $\displaystyle \alpha$ is equal to \_\_\_\_.
Official answer
160
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JEE Main 2021 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.