Question

Find the sum of `1^2 - 2^2 + 3^2 - 4^2 + 5^2 - 6^2........`? ( Please find out the #n^(th) term and then use the sigma method)

Answer 1

The sum is `=(-1)^(n+1)(n(n+1))/2`

Explanation:

The `nth` term is `=(-1)^(n+1)n^2`

The sum is

`S=1^2-2^2+3^2-4^2+5^2-6^2+........+(n-1)^2-n^2`, `AA n in NN`

If `n` is even

`S=(1^2-2^2)+(3^2-4^2)+(5^2-6^2)+......+((n-1)^2-n^2)`

`S=(1-2)(1+2)+(3-4)(3+4)+(5-6)(5+6)+...+((n-1-n)(n-1+n)`

`S=(-1)(1+2)+(-1)(3+4)+(-1)(5+6)+....+((-1)(n-1+n)`

`S=(-1)((1+2)+(3+4)+(5+6)+....+(n-1)+n)`

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

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

If `n` is odd

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

Answer 2

`S_N = -((-1)^N N(N+1))/2`

Note that this is completely general, irrespective of whether the `N`th term is even or odd. An evaluation example can be seen at the bottom.

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DERIVATION

If all the terms were adding, the sum would be:

`sum_(n=1)^(N) n^2 = 1^2 + 2^2 + . . . + N^2`

Since the series is alternating, we can write the sum to include a `(-1)^(n)`:

`sum_(n=1)^(N) (-1)^(n+1) n^2`

`= (-1)^(2)(1)^2 + (-1)^3(2)^2 + (-1)^4(3)^2 + . . .`

`= 1^2 - 2^2 + 3^2 - . . .`

The `N`th term would be given by `(-1)^(N+1)N^2`, and the finite sum at the `N`th term would be found as follows. If this series were not alternating, the sum would have been:

`S = (N(N+1))/2`.

But it's not that. We would have to account for the fact that each term alternates sign, and that means the `(N+1)`th term would be opposite in sign to the `N`th term.

• If we suppose the `N`th term is positive, then it is an odd term. That means to change its sign, we need to multiply by `(-1)^(N)`, which is guaranteed to be a `-1` multiplier.
• The `(N+1)`th term is therefore negative, and an even term. It is not affected by `(-1)^(N+1)` since `N+1` is even.

Lastly, we can rewrite our infinite sum to realize...

`sum_(n=1)^(N) (-1)^(n+1) n^2`

`= ul(-sum_(n=1)^(N) (-1)^(n) n^2)`

...that a negative sign can be factored out of everything. Thus, since `(-1)(-1)^n` is essentially independent of the term index, the finite sum up to the `bbN`th term, defined as `S_N`, is:

`color(blue)(S_N = -((-1)^N N(N+1))/2)`

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EXAMPLE

For example, the sum up to the 3rd term is

`S_3 = -((-1)^3 3(3+1))/2`

`= 6`

And we can easily check this.

`S_3 = 1^2 - 2^2 + 3^2`

`= 1 - 4 + 9 = 6` `color(blue)(sqrt"")`.

For an even `N` example, consider the 6th term I suppose.

`S_6 = -((-1)^6 6(6+1))/2`

`= -21`

And we check to see that...

`S_6 = 1^2 - 2^2 + 3^2 - 4^2 + 5^2 - 6^2`

`= 1 - 4 + 9 - 16 + 25 - 36 = -21` `color(blue)(sqrt"")`

Answer 3

The `n^(th)` term is given by:

`u_n = (-1)^(r+1) r^2`

The sum of the first `n` term is given by:

`sum_(r=1)^n (-1)^(r+1) r^2 =(-1)^(n+1) \ 1/2n(n+1)`

Explanation:

We seek

`S_o = 1^2 -2^2 + 3^2 -4^2 + ... +n^2 \ \ \ \` if `n` odd
`S_e = 1^2 -2^2 + 3^2 -4^2 + ... -n^2 \ \ \ \` if `n` even

Or in Sigma notation:

`S = sum_(r=1)^n (-1)^(r+1) r^2 \ \ \`

We can use the following relationship:

`-r^2 + (r+1)^2 = -r^2 + r^2+2r+1`
`" " = 2r+1`
`" " = r + (r+1)`

Hence we can write :

`-r^2 + (r+1)^2 -= r+ (r+1)` ..... [A]

Similarly we can write:

`r^2 - (r+1)^2 -= -(r+ (r+1))` ..... [B]

Along with the standard summation result:

`S = sum_(r=1)^n r= 1/2n(n+1)` ..... [C]

Using the above relationship we can group terms as follows:

Case 1 : `n` odd

`S_o = 1^2 -2^2 + 3^2 -4^2 + ... -(n-1)^2 + n^2`
`\ \ \ \ = (1^2) + (-2^2 + 3^2) + ( -4^2 + 5^2) + (-(n-1)^2 + n^2)`

Using [A] this becomes:

`S_o = (1) + (2+3) + (4+5) + ((n-1)+n)`
`\ \ \ \ = 1+2+3 + ...+n`
`\ \ \ \ = 1/2n(n+1) \ \ \` using [C]

Case 2 : `n` even

`S_e = 1^2 -2^2 + 3^2 -4^2 + ... +(n-1)^2 - n^2`
`\ \ \ \ = (1^2 -2^2) + (3^2 -4^2) + ... +((n-1)^2 - n^2)`

Using [B] this becomes:

`S_e = (-(1+2)) + (-(3+4)) + ... +(-((n-1)+n))`
`\ \ \ \ = -(1+2+3 ... + n)`
`\ \ \ \ = -1/2n(n+1) \ \ \` using [C]

In summary we have:

`S_o = \ \ \ \ \1/2n(n+1) \ \ \ \` if `n` is odd
`S_e = -1/2n(n+1) \ \ \ \` if `n` is even

And so can readily combine these results to get the general formula:

`sum_(r=1)^n (-1)^(r+1) r^2 =(-1)^(n+1) \ 1/2n(n+1)`

Answer 4

See below.

Explanation:

Calling

`S_o = sum_(k=1)^n(2k-1)^2` and

`S_e = sum_(k=1)^n(2k)^2`

for `n` even we have

`S_(2n) = S_o + S_e = sum_(k=1)^n (2k-1)^2-(2k)^2 =`

`= sum_(k=1)^n(2k-1+2k)(2k-1-2k) = -sum_(k=1)^n (4k-1) =`

`= n-4 sum_(k=1)^nk = n-4((n+1)n)/2 = n-2n(n+1) = -n(2n+1)`

and for `n` odd

`S_(2n+1)=S_(2n)+(2 (n + 1) - 1)^2 = (n+1)(2n+1)`

Resuming

`{(S_(2n) = -n(2n+1)),(S_(2n+1)= (n+1)(2n+1)):}`