Question

Need 1 help
Plz help me........

Answer 1

what is it

Answer 2

If \[1^{2}+2^{2}+3^{2}+.......+512^{2}=m\]\[how is 2^{2}+4^{2}+6^{2}........+1024^{2}=4m\]
Plz explain

Answer 3

you can use the formula for the sum of squares i believe
do you know it?

Answer 4

no bro @satellite73

Answer 5

i.e.
\[\sum_{k=1}^n k^2=\frac{n(n+1)(2n+1)}{6}\]

Answer 6

Or straight into the fire :P
\[n=2^2 + 4^2 + 6^2 +...1024^2\\=(2\cdot 1)^2 +(2\cdot2)^2 + (2\cdot 3)^2+...(2\cdot512)^2\\=4\cdot1^2 + 4\cdot2^2+4\cdot3^2 + ...4\cdot512^2\]

And you get the idea, hopefully :D

Answer 7

so
\[\sum_{k=1}^{512}=\frac{512(513)(1025)}{6}=m\]

Answer 8

actually that is a much better idea

Answer 9

you and your formulas XD
And I thought I was the one with that tendency ^_^

Answer 10

\[\sum_{k=1}^n(2k)^2=4\sum_{k=1}^nk^2\]

Answer 11

^And THAT is much neater ^_^

Answer 12

i was getting there, just a little slower than necessary

Answer 13

@terenzreignz great approach nw taking common 2^2 from each term we can say the expression is =4m

Answer 14

Yup. Just factor out the 4, and you'll have the m left :D

Answer 15

u too helped me a lot bro @satellite73

Answer 16

yup i have got it nw bro thnq very much @terenzreignz

Answer 17

That's all I need to know ^_^

Answer 18

@satellite73 thnx to u too bro