If 12020-22019-32018-20201-2020 - 2021 - k- then the value of k isA- 337B- 2019C- 365D- 2019-2

The correct option is A 337From the given series, r^th term is nbsp;t_r=r(2021 r) Thus, S=∑_r=1^2020 r(2021 r) =2021 ∑_r=1^2020r ∑_r=1^2020r^2 =2021 ·2020 × ...

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Byju's Answer

Standard XII

Mathematics

Summation by Sigma Method

If 12020+2201...

Question

If $(1) (2020) + (2) (2019) + (3) (2018) + \dots + (2020) (1) = 2020 \times 2021 \times k,$ then the value of $k$ is

365

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337

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2019

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\frac{2019}{2}

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Solution

The correct option is B $337$
From the given series,
$r^{t h}$ term is $t_{r} = r (2021 - r)$

Thus, $S = 2020 \sum r = 1 r (2021 - r)$
$= 2021 2020 \sum r = 1 r - 2020 \sum r = 1 r^{2}$
$= 2021 \cdot \frac{2020 \times 2021}{2} - \frac{2020 \times 2021 \times 4041}{6}$
$= \frac{2021 \times 2020}{2} (2021 - \frac{4041}{3})$
$= 2021 \times 2020 \times 337$
$∴ k = 337$

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If (1)(2020)+(2)(2019)+(3)(2018)+⋯+(2020)(1)=2020×2021×k, then the value of k is

The correct option is B 337From the given series, rth term is tr=r(2021−r) Thus, S=2020∑r=1r(2021−r) =20212020∑r=1r−2020∑r=1r2 =2021⋅2020×20212−2020×2021×40416 =2021×20202(2021−40413) =2021×2020×337 ∴k=337

If $(1) (2020) + (2) (2019) + (3) (2018) + \dots + (2020) (1) = 2020 \times 2021 \times k,$ then the value of $k$ is