Q.

The value of the sum 1.22+2.32+3.42+…+19⋅202 is

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a

44100

b

44000

c

41230

d

40000

answer is C.

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Detailed Solution

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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