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 The value of the sum 1.22+2.32+3.42++19202 is 

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a
44100
b
44000
c
41230
d
40000

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detailed solution

Correct option is C

Given sum=∑i=220 (r−1)r2=∑i=120 r3−∑i=120 r2=20212=44100−2870=412302-2021416    ∑i=1n r3=nn+122  , and ∑i=1n r2 =nn+12n+16


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Let m be a positive integer, then S=k=1mk1k+1k+1+1k+2++1m is equal to :


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