#summation
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limber plover
First we distribute the n to 2n+1 which will give us 2n^2+n we can now rewrite the equation as sum n=1 to 10 of 2n^2 + sum n=1 to 10 of n which if we solve the 2n^2 sum. We should know that the sum n=1 to k of n^2 is equal to
(k(k+1)(2k+1))/6 and if we have 2n^2 we should get (k(k+1)(2k+1))/3 where we just plug in k in this case k=10 that sum will give us 770 . For the second sum we get sum n=1 to 10 of n is going to be (k(k+1))/2 again, k=10 this will give us 55 and we just add up the two results giving us our solution 770+55=825
dull stone
got it! thank youu
How do we derive the formulas?? Like what if we have greater powers
how do we deal with that
limber plover
well for higher degrees it just gets more bigger and it is hardly used in general practice for example if we have a power of 6 we get
(k(k+1)(2k+1)(3k^4+6k^3-3k+1))/42
dull stone
ohh okayy
quasi mural
A method to determine higher degrees is to use telescopic sums :
For example let Sn be the the sum of k^3 if you want to determine the sum of k^3 let’s say well for any k you have (k+1)^4-k^4=4k^3+6k^2+4k+1 if you sum till n on each side you’ll see that the sum on the left is (n+1)^4-1 (it telescopes)and on the right you’ll have 4Sn+n(n+1)(2n+1)+2n(n+1)+now you can find Sn by solving an equation so basically if you know the formula for any degree that is inferior you can determine the formula for that degree with this