Sequence and series

Question

Sum to n terms of series 

 

1^2 -  2^2 + 3^2 - 4^2 + 5^2 - 6^2 +....... 

Answer 1

Hi Sam,

 

The hardest part of a series is finding the general form.

 

1) this is definitely a series of the type ∑i²

2) but we also need to consider the fact that it's an alternating series.

 

So we would have

n

∑    (-1)⁽ⁱ⁺¹⁾ i²

i=1

 

The (-1)^(i-1) gives us the alternating form.  Using i+1 as the power makes sure the odd numbers are added and the even ones are subtracted.

 

Now the sum of an i² series is n(n+1)(2n+1)/6 but we have an alternating series so we need to look further.

 

We can look at this series as 

 

n/2                  n/2

∑     (2i-1)² -     ∑   (2i)²

i=1                 i=1

 

Where the first term give us the positive squares of the odd integers and the second term gives us the negative squares of the even integers.  Notice I changed the series so we only go to n/2 since I'm multiplying i by 2 in both series.

 

We can rewrite this as

 

n/2             n/2
∑ 4i²-4i+1 - ∑ 4i²
i=1             i=1

 

To make my typing a bit less frustrating I have left off the "from i=1 to n/2" from the summations below but they should be there.
∑ 4i²-∑4i+∑1 - ∑ 4i²

 

This reduces to

∑1 - ∑4i

 

which is equal to 

∑1 - 4∑i

 

Now ∑i = (n/2)(n/2+1)/2 so 4∑i = n(n/2+1)

and ∑1 = n/2

 

So what we're left with is

n/2 - n(n/2+1)

 

Answer 2

it's

 

(1-2)(1+2)+(3-4)(3+4)+(5-6)(5+6)+...

 

or

 

-3-7-11-15-...

 

the first term is -3, difference is -4\

 

apply the formula for arithmetic series