Question

how to find sum of series 6, 12, 20, 30, 42 .... upto 211 terms

Answer 1

Just keep summing them up!
LOL

Answer 2

just kidding!
There is a general formula for terms which have constant increase, but you don't have that case here. wait for a min and I'll see if I could think of something...

Answer 3

@smarty20two I am thinking of writing an equation for this series. They all lie on a line y = mx + 4 where m =2

Answer 4

good note! that would be an excellent start!
since we have the equation, we can probably write it as a sum

Answer 5

@smarty20two can we find integration of this equation to find sum ? Is integration applicable for discreet values ? IDK :-|

Answer 6

That is a question that I am not certain of, even though integration is applied to find error estimation, I still have not understood why we are considering continuous areas to be relevant to discrete calculations

Answer 7

your answer would be 45576
Don't ask me how!

Answer 8

I can't provide you with a satisfactory solution, but here's the idea.
Since aporvearyan stated that they all lie on the same line (and he was correct), we can write the equation, y=2x+4. from here, if you use the summation rule, we have:
\[\sum_{x=1}^{211} 2x+4=2\sum_{x=1}^{211}x+\sum_{x=1}^{211} 4=2\frac{ (1+211)(211) }{ 2 }+211\times4=45576\]

Answer 9

satisfied with the answer?

Answer 10

I told ya you wouldn't like seeing it! LOL

Answer 11

@smarty20two yup .... nice one ... lol

Answer 12

Thanks for the idea that they were on a straight line. I doubt that I could have solved it without that equation. However, you should notice that in case they give you some equation of degree 2 or higher or even worse, numbers with oscillating patterns, then these simple logics won't come handy!

Answer 13

@smarty20two I don't think this is write answer. I checked it on python interpreter ... But I am not sure if my code was write

Answer 14

@smarty20two do you know any programming language ?

Answer 15

not for the time being... you can ask me the same question in 6 months and my answer will be different for sure

Answer 16

tell me, is programming difficult?

Answer 17

@smarty20two Programming is very easy but logical.

Answer 18

Then I shouldn't face much trouble in it! Thanks for relieving me!

Answer 19

Hey, I though I answered that!

Answer 20

@smarty20two the answer is wrong my friend ... but we will find it

Answer 21

yes, because the equation of the line is incorrect
I took the equation without analyzing it

Answer 22

they're not on the same line!

Answer 23

at least not as far as I can see it

Answer 24

@smarty20two yes the series lies on parabola not on straight line :( I got it wrong Sorry :'(

Answer 25

then that would change everything as I have said before

Answer 26

@smarty20two yes, let me try to find new equation .... then your amazing method would work :)

Answer 27

yeah, you find an equation, I'll solve the question. good team work

Answer 28

I think the answer for this question is something like this:
a1 = 6
a2 = a1+2(3)
a3 = a2 + 2(4)
...
an = \[\sum_{1}^{n+1}2k\]
Therefore, an = (n+1)(n+2) = n^2 + 3k + 2, this formula gives you the nth term in the sequence you mentioned above.

For its sum you need to add from k = 1 to k = n, (k+1)(K+2)... sorry this is my first time posting here and I do not know how to get the sigma to work.

Your answer for the sum of first n terms = \[\sum_{1}^{n}k^2\] + \[\sum_{1}^{n}3k\] + \[\sum_{1}^{n}2\](n/6)*(n+1)*(2n+1) + 3(n^2+n)/2 + 2n.

So the sum of first 211 terms in the sequence is : (211/6)*(211+1)*(2(211)+1) + 3(211^2+211)/2 + 2(211) = 3,221,126