For two non-zero numbers m and n define * by: \sum_{k=1}^{2018}=(1/k^2 + k) =

TS ICET 23 May 2018 Shift 2 Paper

Question : 74 of 200

Marks: +1, -0

For two non-zero numbers m and n define * by:

∑↙{k=1}↖{2018}=(1/k^2 + k)

Solution:

1/k^2×k=k+1

Σ_{k=1}^{2018} \;{1}/{k^{2}} × k=∑_{k=1}^{2018}(k+1)

= 2 + 3 + …. 2019

= (2019)(1010)-1

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