Question

Evaluate
1 + (1+2) + (1+2+3) + (1 +2 + 3+ 4) + … upto n terms.

Answer 1

@ajprincess

Answer 2

Am really sorry. I learnt series long ago.Hwever much i try I dnt seem to remember it:(

Answer 3

Can you tag any expert in my question please

Answer 4

Who is free now

Answer 5

ya sure:)
@ganeshie8, @amistre64, @aravindG pls help:)

Answer 6

\[\sum\sum n\]maybe?

Answer 7

Notice that the nth term will be sigma n

Answer 8

You can write the sum as the following:
\[\frac{1(1+1)}{2}+\frac{2(2+1)}{2}+\frac{3(3+1)}{2}+\frac{4(4+1)}{2}+\cdots\]
The \(n-\)th term would then be \(\dfrac{n(n+1)}{2}\).

Answer 9

So far I am at

\[S=\frac{ 1 }{ 2 }(1(1+1)+2(2+1)13(3+1)+...+n(n+1))\]
looks like

\[S= \frac{ 1 }{ 2 } \sum_1^n n(n+1)\]

Answer 10

yep^^

Answer 11

Can we use @amistre64 's profile pic her.

I mean can we use integration as the sum of the series here ?

Answer 12

0+1 = 1
1+2 = 3
3+3 = 6
6 + 4 = 10
10 + 5 = 15

1 3 6 10 15
2 3 4 5
1 1 1
\[1+2n+\frac12n(n-1)~:~n=0,1,2,...\]

Answer 13

the series is discrete tho, not continuous

Answer 14

Ok thank you all of you .

Answer 15

\[S_n= \frac{ 1 }{ 12 }\left( n(n+1)(2n+1)+3n(n+1) \right)\]

Is what I got from evaluating my stuff above.

Answer 16

I only checked for the first few terms though, there it seems to hold true.

Answer 17

\[S_n=1+2n+\frac12n(n-1)~:~n=0,1,2,...\]
\[S_n=\frac{2+4n+n^2-n}{2}\]
\[S_n=\frac{n^2+3n+2}{2}; would have to let n=n+1 to match siths setup\]

Answer 18

@Spacelimbus, You can probably prove it by induction.

Answer 19

are you all satisfied with the discussion .

I am fully satisfied so closing the question . Thank you sir and madam.