Alternating Sum Of Squares

We can group the given expression into:

$(40^2 - 39^2) + (38^2 - 37^2) + ... + (2^2 - 1^2)$ , and with the use of the identity $a^2 - b^2 = (a+b)(a-b)$

$= (40^2 - 39^2) + (38^2 - 37^2) + ... + (2^2 - 1^2)$ $= (40+39)(40-39) + (38+37)(38-37) + ... + (2+1)(2-1)$ $= (40+39)(1) + (38+37)(1) + ... + (2+1)(1)$ $= (40 + 39 + 38 + 37 + ... + 2 + 1) = (40)(41)/2 = 820$

Answer: $820$

If we look at the differences on a pairwise basis, we can can rewrite the problem as $\sum_{n=1}^{20} (2n)^2 - (2n - 1)^2$ or as the recurrence: $\begin{aligned} f(0) &= 0 \\ f(n) &= f(n-1) + (2n)^2 - (2n - 1)^2, \; 1 \leq n \leq 20 \end{aligned}$ Expanding the first term in the pairwise difference simplifies the recurrence relation: $\begin{aligned} f(n) &= f(n-1) + (2n)^2 - (2n - 1)^2 \\ &= f(n-1) + 4n^2 - 4n^2 + 4n - 1 \\ &= f(n-1) + 4n - 1 \end{aligned}$ When a recurrence relation is of the form $f(n-1) + g(n)$ where $g(n)$ is a polynomial in $n$ , we can write a generic function of one degree higher than $g$ to find a closed form solution to $f$ : $f(n) = An^2 + Bn + C$ Since we have 3 unknowns, we need 3 equations. Choosing $n=1,\,2,\,3$ , we have: $\begin{aligned} A + B + C &= 3 \\ 4A + 2B + C &= 10 \\ 9A + 3B + C &= 21 \end{aligned}$ Solving this system reveals $A=2, \, B=1$ and $C=0$ , which gives the closed form solution: $f(n) = 2n^2 + n$ Evaluating for $n=20$ we find that $f(20) = 2 \times 400 + 20 = 820$ .

Sum of squares of first n natural numbers = [ n (n+1) (2n+1) ] / 6

Sum of squares of first n odd natural numbers = [ n (2n-1) (2n+1) ] / 3

(refer Proof at http://www.9math.com/book/sum-squares-first-n-odd-natural-numbers)

The expression in the question equals to

( Sum of squares of first 40 natural numbers - Sum of squares of first 20 odd natural numbers) - Sum of squares of first 20 odd natural numbers

= Sum of squares of first 40 natural numbers - 2 x Sum of squares of first 20 odd natural numbers

= [ 40 (40+1) (2x40+1) ] / 6 - 2 [ 20 (2x20-1) (2x20+1) ] / 3

= 820

$(40^2-39^2)+(38^2-37^2)+...+(2^2-1^2)$

$=(40+39)+(38+37)+...+(2+1)$

$=3+7+...+79$

$=\frac{20}{2}\times(3+79)$ ; $[3+(n-1)4=79\Leftrightarrow n=20]$

$=820$

(2..40).step(2).to a.collect{|i| i*i}.inject(:+)-(1..39).step(2).to a.collect{|i| i*i}.inject(:+)

Let's rewrite...

$(40^2+38^2+...+4^2+2^2)-(39^2+37^2+...+3^2+1^2)$

$=(40^2-39^2)+(38^2-37^2)+...+(4^2-3^2)+(2^2-1^2)$

$=(40+39)(40-39)+(38+37)(38-37)+...+(4+3)(4-3)+(2+1)(2-1)$

$=(40+39)+(38+37)+...(4+3)+(2+1)$

$=40+39+38+37+...+4+3+2+1$

Now it's just a simple arithmetic progression... It's the sum of first 40 positive integers...

We know the formula for finding out the sum of first $n$ positive integers...

That is, $\frac {n(n+1)}{2}$

Simple plugging in $n=40$ , we get...

$\frac {40 \times (40+1)}{2} = \frac {40 \times 41}{2} = 20 \times 41 = 820$

Hence, the required answer is $\fbox {820}$

Find the difference of 40 squared and then 39 squared and do the same for 38 squared and 37 squared. Compare both and u will realise as the number go down they reduce by 4. So take 79 and keep minus with 4 till you reach 3 because the last one is 2 squared minus 1 squared which is 3.

The question can be broken into two series:

First series is the sum of squares of first 20 even natural numbers: $(40^2 + 38^2 + ... + 4^2 + 2^2)$

Second series is the sum of squares of first 20 odd natural numbers: $(39^2 + 37^2 + ... + 3^2 + 1^2)$

The answer is the difference between these two sums.

Therefore,

$(40^2 + 38^2 + ... + 4^2 + 2^2)$ $= \frac{2n(n + 1)(2n + 1)}{3}$ $= \frac{2 . 20 . (20 + 1)(2 . 20 + 1)}{3} = 11480$

and

$(39^2 + 37^2 + ... + 3^2 + 1^2)$ $= \frac{n(2n - 1) (2n + 1)}{3}$ $= \frac{20 . (2 . 20 - 1) (2 . 20 + 1)}{3} = 10660$

Hence, the final solution,

$(40^2 + 38^2 + ... + 4^2 + 2^2) - (39^2 + 37^2 + ... + 3^2 + 1^2)$ $= 11480 - 10660$ $=820$ which is the required answer.

( $40^{2}$ - $39^{2}$ )+( $38^{2}$ - $37^{2}$ )+...+( $2^{2}$ - $1^{2}$ )=(40+39)(40-39)+(38+37)(38-37)+...(2+1)(2-1)=40+39+38+37+...+2+1=820

40^2-39^2 +38^2-37^2............. 2^2-1^2 (40+39)(40-39)+(38+37)(38-37)..............................(2+1)(2-1) Taking one common , we get 1* (1+2+3+4+5........................40) 1*(40(40+1)/2 = 820

I know that this isn't the correct way, but I used the following python code:

list_1=[x**2 for x in range(2,41,2)]
list_2=[x**2 for x in range(1,40,2)]
print sum(list_1)-sum(list_2)

On opening the brackets and rearranging the given sequence it follows -

(40^2 - 39^2) + (38^2 - 37^2) + (36^2 - 35^2) + ......... +(2^2 - 1^2) this implies, (40+39)(40-39) + (38+37)(38-37) + ........... +(2+1)(2-1) this implies, (40+39)(1) + (38+37)(1) + ....... +(2+1)(1) this implies, 40+39+38+37+.......+2+1

(40^2-39^2)+(38^2-37^2)+...+(2^2-1^2)=

(40+39)(40-39)+(38+37)(38-37)+...+(2+1)(2-1)=

79+75+...+3=

(20/2)(79+3)=(10)(82)=820

(40^2 + 38^2 +…+ 4^2 + 2^2) − (39^2 + 37^2+…+3^2 + 1^2) = (40^2 - 39^2) + (38^2 - 37^2) +...+(4^2 - 3^2) + (2^2 - 1^2) = 79 + 75 + 71 + 67 + ....+ 11 + 7 + 3 = 20/2 (79 + 3) = 10 x 82 = 820