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Sum the series 1+(1+2)+(1+2+3)+...n terms.

what is the n term???
not given
so where would it end???????? i mean what would be the last no
dont knw dis is d ques. which i hve copied from d book.


N th term is =n(n+1)/2=n²+n/2=n²/2+n/2


after simplyfing it equals =n(n+1)(n+2)/6
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In the series 1+(1+2)+(1+2+3)....., the nth term is the sum of numbers from 1 to n.
So the nth term is n(n+1)/2.
So we need to find the sum of a series whose nth term is n(n+1)/2.

\Sigma  \frac{n^2}{2}= \frac{n(n+1)(2n+1)}{6*2}=\frac{n(n+1)(2n+1)}{12}  \\  \\ \Sigma  \frac{n}{2}= \frac{n(n+1)}{2*2} =\frac{n(n+1)}{4} \\  \\  \Sigma  \frac{n(n+1)}{2} = \Sigma  \frac{n^2+n}{2}= \Sigma  \frac{n^2}{2}+ \Sigma  \frac{n}{2}  \\  \\ \frac{n(n+1)(2n+1)}{12}+\frac{n(n+1)}{4}  \\  \\ \frac{n(n+1)}{12}((2n+1)+3)=\frac{n(n+1)}{12} (2n+4) \\  \\ \frac{n(n+1)(n+2)}{6}
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any constan multiplication shuld be outside the sigma
which constant?
ur answer is correct but ur notation is wrong
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