An algebra problem by kritarth lohomi
n = 1 ∑ 1 0 0 n n + n 2 = ?
The answer is 5150.
This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try
refreshing the page, (b) enabling javascript if it is disabled on your browser and,
finally, (c)
loading the
non-javascript version of this page
. We're sorry about the hassle.
2 solutions
Moderator note:
Yes correct. Alternatively we could have n = 1 ∑ 1 0 0 ( n + 1 ) = 2 + 3 + 4 + … + 1 0 1 = 2 1 0 1 × 1 0 2 − 1 .
C o r r e c t ! !
I did it mentally!
Log in to reply
@Anik Mandal Me too! Hi5!
actually, I too did it mentally
I did mentally too. I have practiced to do simple summatories like this one.
I did it in same way!
hmm, ....... right ..!
Summation of (n+n^2)/n = 1+n....then putting n=1...to 100...we obtain an A.P. series 2+3+4+5+........+101. Sn= (100)/2{2x2 + (100-1)1} = 50x103 = 5150 (ans)
Moderator note:
Correct. Treating it as a sum of an arithmetic progression is the same thing. The benefit for this method is that you don't have to remember the identity k = 1 ∑ n k = 2 n ( n + 1 ) .
n = 1 ∑ 1 0 0 n n 2 + n = n = 1 ∑ 1 0 0 n n ( n + 1 ) = n = 1 ∑ 1 0 0 ( n + 1 )
= n = 1 ∑ 1 0 0 n + 1 0 0
= 2 1 0 0 ( 1 0 0 + 1 ) + 1 0 0
= 5 0 × 1 0 1 + 1 0 0
= 5 0 5 0 + 1 0 0
= 5 1 5 0