Question

My question is attached.

Answer 1

Question D

Answer 2

did you show question b) c ) try to use them !

Answer 3

I have tried.

Answer 4

ok what make a problem to you in the expression given in c) !

Answer 5

don't forget ! you're trying to find \[\sum_{1}^{n} i^2\]
so all you have to know is \[\sum_{1}^{n}i=n(n+1)/2\]

Answer 6

did you get it .?!

Answer 7

\[3\sum_{1}^{n} i^{2}+3\sum_{n}^{1}i+n=(n+1)^{3}-1\]

Answer 8

yeaah ! and as I told you before \[\sum_{1}^{n}i=n(n+1)/2\]

Answer 9

\[3\sum_{1}^{n}i ^{2}+3n(n+1)/2+n=n ^{3}+3n ^{2}+3n\]

Answer 10

This is where I get stuck.

Answer 11

put 3n(n+1)/2 in the other side ! (ithink you know how to do it ? right :) :) )

Answer 12

oh sugar.

Answer 13

yeaaah !

Answer 14

I just got it

Answer 15

That's my problem, PATIENCE

Answer 16

you're really smart !---this problem means a lot to me :) :) !

Answer 17

you're really smart !---this problem means a lot to me :) :) !

Answer 18

is that a bug or did you copy it ?!

Answer 19

I copied,and actually I think I spoke to soon looool.

Answer 20

what do you get after you subtract from both sides?

Answer 21

something very messy my friend

Answer 22

can I multiply by 2?

Answer 23

ok ! \[3\sum_{1}^{n}i^2 = n^3+3n(n+1)-3n(n+1)/2=n^3+3n(n+1)/2=n(3(n+1)/2+n^2)\] ANd yes you can !

Answer 24

LHS=n^3+n?????

Answer 25

I factorized by n !

Answer 26

sorry ! i got to go !

Answer 27

thanks for your help.