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
The sum is
Explanation:
The
The sum is
If
If
S_N = -((-1)^N N(N+1))/2
Note that this is completely general, irrespective of whether the
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
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
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
- 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) sinceN+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
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
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"")
The
u_n = (-1)^(r+1) r^2
The sum of the first
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 \ \ \ \ ifn odd
S_e = 1^2 -2^2 + 3^2 -4^2 + ... -n^2 \ \ \ \ ifn 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 :
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 :
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) \ \ \ \ ifn is odd
S_e = -1/2n(n+1) \ \ \ \ ifn 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)
See below.
Explanation:
Calling
for
and for
Resuming