prove by induction

[chipatel87]

1 − 4 + 9 − 16+· · ·+(−1)^n−1n^2 = (−1)^n−1(1 + 2 + 3+· · ·+n)

step 1: 1 =1
step 2: (−1)^k−1k^2
step 3: k +1

((−1)^n−1n^2 )+1 is this correct

 

[Deleted member 4993]

1 − 4 + 9 − 16+· · ·+(−1)^n−1n^2 = (−1)^n−1(1 + 2 + 3+· · ·+n)

step 1: 1 =1
step 2: (−1)^k−1k^2
step 3: k +1

((−1)^n−1n^2 )+1 is this correct

Is that expression correct? From regular interpretation, it does not seem so:

for n = 2

(−1)^n−1n^2 = (-1)² - 1*2² = -1 - 4 = -5 (not -4 as shown on the left-hand-side)

 

[chipatel87]

ok correction

1 = 1
1 - 4 = - (1 + 2)
1 - 4 + 9 = (1 + 2 + 3)
1 - 4 + 9 – 16 = - (1 + 2 + 3 + 4)
1 - 4 + 9 – 16 + 25 = (1 + 2 + 3 + 4 + 5)
Guess a general formula and prove it by mathematical induction
that is the question

forrmula for left side is (-1)^(n+1) (((n*(n+1))/2) but i am having problem with the right one's formula and how to prove it by induction

 

[Deleted member 4993]

1 = 1
1 - 4 = - (1 + 2)
1 - 4 + 9 = (1 + 2 + 3)
1 - 4 + 9 – 16 = - (1 + 2 + 3 + 4)
1 - 4 + 9 – 16 + 25 = (1 + 2 + 3 + 4 + 5)
Guess a general formula and prove it by mathematical induction
that is the question

forrmula for left side is (-1)^(n+1) (((n*(n+1))/2) but i am having problem with the right one's formula and how to prove it by induction

So the sequence goes like this:

1-4+9-16.....+(-1)ⁿ⁺¹*n² + (-1)ⁿ⁺²*(n+1)²..... = (-1)ⁿ⁺¹[n²/2 + n/2] + (-1)ⁿ⁺²*(n+1)² = (-1)ⁿ⁺²[-n²/2 - n/2] + (-1)ⁿ⁺²*(n+1)² = (-1)ⁿ⁺²[-n²/2 - n/2 + (n+1)² ]

= (-1)ⁿ⁺²[(n+1){n + 1 - n/2}] = (-1)ⁿ⁺²[(n+1){n + 2}/2]

 

[chipatel87]

k+1

i am still confuce about the last step.

solve for k+1: (-1)^(k+2) *(k+1)(k+2)/2
(-1)^(k) (-1)^2*(k+1)(k+2)/2
( (-1)^k(k+1)(k+2))/2

is this correct for the third step? am i done? do i need to add anything else