# 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 ** 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

- 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 **finite sum up to the** **th term**, defined as

#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 \ \ \ \ # 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 :** **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 :** **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) #

See below.

#### Explanation:

Calling

for

and for

Resuming