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)

4 Answers
Aug 18, 2017

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

Aug 18, 2017

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

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


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 Nth term would be given by (-1)^(N+1)N^2, and the finite sum at the Nth 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 Nth term.

  • If we suppose the Nth 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 bbNth term, defined as S_N, is:

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


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"")

Aug 18, 2017

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)

Aug 18, 2017

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)):}