Question

Prove that for every positive integer n
1^2 -2^2+3^3-...+(-1)^(n-1) n^2 = (-1)^(n-1) (n(n+1)/2)

Answer 1

by induction:suppose it is true for n then try to prove it true for n+1

Answer 2

Okay then I get (-1)^n ((n+1)(n+2))/2

Answer 3

how do you know it's true though?

Answer 4

try it first for n=1,2,3...

Answer 5

okay but how do you show it for any n

Answer 6

S(n)=1^2 -2^2+3^2-...+(-1)^(n-1) n^2 = (-1)^(n-1) (n(n+1)/2)

we want to prove:

S(n+1)=(-1)^n ((n+1)(n+2))/2

but:

S(n+1)=S(n)+(-1)^(n)( n+1)^2

so S(n+1)=(-1)^(n-1) (n(n+1)/2)+(-1)^(n)( n+1)^2

now factor (-1)^(n-1)*(n+1) out:

S(n+1)=(-1)^(n-1)*(n+1)[n/2-n-1]

=(-1)^(n-1)*(n+1)(-n-2)/2

and since (-1)^(n-1)=(-1)^n*(-1)^(-1)=-(-1)^n

then S(n+1)=(-1)^(n)*(n+1)(n+2)/2 which is what we wanted to prove

Answer 7

sorry but did you get the s(n) in S(n+1)=S(n)+(-1)^(n)( n+1)^2?

Answer 8

but where did you**

Answer 9

each sum is the previous one + the last term((-1)^(n-1) n^2) which depends on n..

Answer 10

in our case i just put for the last term n+1 instead of each n

Answer 11

Oh I see thank you that make sense

Answer 12

welcome

Answer 13

sorry I have one more question about this I just don't see how S(n) which is the sum of all the previous term can be equal to just the previous term (-1)^(n-1) (n(n+1)/2)?

Answer 14

is (-1)^(n-1) (n(n+1)/2) not just giving us 1 term

Answer 15

one in my sentence meant previous sum...not term

Answer 16

OH I SEE

Answer 17

you took the sum which it was equal to before n+1

Answer 18

ohhh I see okay okay thanks a lot haha sorry to bug you

Answer 19

nope