Question

Find the result of :
1^3 - 2^3 + 3^3 - 4^3 + ... + 49^3 - 50^3

Answer 1

do u have formula or property or something to simplify it

Answer 2

1^3 - 2^3 + 3^3 - 4^3 + ... + 49^3 - 50^3
=(1-2)(1^2 + 1*2+2^2) + (3-4) (3^2 + 3*4 + 4^2) + ... + (49-50)(49^2 + 49*50+50^2)
= - (1^2 + 2^2 + 3^2 + 4^2 +... + 50^2) - ( 1*2 + 3*4 + 5*6 + ... + 49*50)

Answer 3

how do u get that, what formula have u used ?

Answer 4

oh, i see.
from. (a-b)^3 = a^3 - 3a^2b + 3ab^2 - b^3
so, a^3 - b^3 = (a-b)^3 + 3ab(a-b) right ??

Answer 5

but i dont see like u done it, what formula is it ?

Answer 6

a^3 - b^3 = (a-b) (a^2 + ab + b^2)

Answer 7

oh yea, you are correct.
next, how do u calculate of 1^2 + 2^2 + 3^2 + 4^2 +... + 50^2 ?
similar question for 1*2 + 3*4 + 5*6 + ... + 49*50) ?

Answer 8

1^2 + 2^2 + 3^2 + 4^2 +... + 50^2
is calculated by sum of n squares .. there is a formula :
n(n+1)(2n+1) / 6
plug in it 50.

Answer 9

ok, thanks..
so, 1^2 + 2^2 + 3^2 + 4^2 +... + 50^2 = 50(51)(101)/6 = 42,925

Answer 10

how about u with 1*2 + 3*4 + 5*6 + ... + 49*50) ?

Answer 11

now for 1*2 + 3*4 + 5*6 + ... + 49*50
we have for the nth term
n(n+1) = n^2 + n
so we have again something like
1^2 + 1 + 2^2 + 2 + 3^2 + 3 + ...
so we have
again sum of n squares but this time the last is 49
and we have sum of n (not squares ) which is given by n(n+1)/2 again last term is 49

Answer 12

so, i see that
1^2 + 1 + 2^2 + 2 + 3^2 + 3 + ...
= (1^2 + 2^2 + 3^2 + ...+ 49^2) + (1+2+3+ ... + 49) right ?

Answer 13

yes

Answer 14

ok, now make sense for me... thank u very much @sauravshakya and @Coolsector

Answer 15

:)