Sum of odd squares

The sum of odd squares up to 2n-1 is equal to the sum of squares up to 2n minus the sum of even squares up to 2n.

We're given the sum of squares up to n as $S_{sq}= \displaystyle \sum_{i=1}^n i^2 = \frac {n(n+1)(2n+1)}{6}$ .

From here, we see the sum of squares up to 2n is $S_{sq2n}=\displaystyle \sum_{i=1}^{2n} i^2 = \frac {2n(2n+1)(4n+1)}{6}$

The sum of even squares is

$2^2 + 4^2 + 6^2 \ldots = (1 \cdot 2)^2 + (2 \cdot 2)^2 + (3 \cdot 2)^2 \ldots$

$=4 \cdot 1^2 + 4 \cdot 2^2 + 4 \cdot 3^2$

$=4 \cdot (1^2 + 2^2 + 3^2)$

$=4 \cdot S_{sq}= 4 \cdot \displaystyle \sum_{i=1}^n i^2 = 4 \cdot \frac {n(n+1)(2n+1)}{6}$

Hence the sum of odd squares is: $\frac {2n(2n+1)(4n+1)}{6} - 4 \cdot \frac {n(n+1)(2n+1)}{6}$

$=\frac{n(2n-1)(2n+1)}{3}$

$=\frac{n(4n^2 -1 )}{3}$

$=\frac{4n^3-n}{3}$

So the coefficient of $n^3$ is 4/3. a=4, b=3, a+b=7.

$\sum_{k=1}^n (2k - 1)^2 = \sum_{k=1}^n (4k^2 - 4k + 1) = 4\sum_{k=1}^n k^2 + \sum_{k=1} (-4k + 1)$ . Only the first term contributes a $n^3$ term. The term is equal to 4 times that of the $n^3$ term in the sum of squares formula $\frac{n(n+1)(2n + 1)}{6}$ . Hence $A = 4 \left(\frac{2}{6}\right) = \frac{4}{3}$ and so $a + b = 4 + 3 = 7$ .

First of all, notice that $1^2+2^2+\cdots+(2n)^2=(1^2+3^2+\cdots+(2n-1)^2)+(2^2+4^2+\cdots+(2n)^2)$ . Now, using the equality $1^2+2^2+\cdots n^2=\dfrac{n(n+1)(2n+1)}{6}$ for both the left hand side and the second monomial on the right hand side we get that $n$ satisfies the equation $\dfrac{2n(2n+1)(4n+1)}{6}=An^3+Bn^2+Cn+D+4\dfrac{n(n+1)(2n+1)}{6}$ . From this last equation, we match the $n^3$ terms and obtain that $\frac83=A+\frac43$ .

Let S=1^(2)+3^(2)+... ... ...+(2n-1)^(2)

     ={1^(2)+2^(2)+3^(2)+4^(2)+... ...+(2n-1)^(2)+(2n)^(2)} - {2^(2)+4^(2 )+         6^(2)+... ...+(2n)^(2)}

     ={(2n)(2n+1)(4n+1)/6} -2^(2){n(n+1)(2n+1)/6}

     ={(4n^(2)+2n)(4n+1)/6} - {(4n^(2)+4n)(2n+1)/6}

     ={16n^(3)+4n^(2)+8n^(2)+2n - 8n^(3) - 4n^(2) - 8n^(2) - 4n}/6}

     ={4n^(3) - n}/3

     ={4n^(3)/3}+0*n^(2) - (n/3)+0

So A=4/3; a=4,b=3 ;a+b=7

Let the general term of the odd squares series be $T_{r}$ = $(2r - 1)^{2}$ = 4 $r^{2}$ - 4 $r$ + 1. We need to sum the series upto ' $n$ ' terms.

$\sum_{r=1}^n$ $T_{r}$ = $\sum_{r=1}^n$ (4 $r^{2}$ - 4 $r$ + 1)

= 4( $\sum_{r=1}^n$ $r^{2}$ ) - 4( $\sum_{r=1}^n$ $r$ ) + ( $\sum_{r=1}^n$ 1)

= $\frac{4}{6}$ $n$ ( $n$ + 1)(2 $n$ + 1) - $\frac{4}{2}$ $n$ ( $n$ + 1) + $n$

We need the coefficient of $n^{3}$ but still we will simplify the expression to see that A , B , C and D are real numbers.

On little simplification we get $\sum_{r=1}^n$ $T_{r}$ = $\frac{4}{3}$ $n^{3}$ - $\frac{n}{3}$ . Now we see that A = $\frac{4}{3}$ = $\frac{a}{b}$ . Since a and b are co-prime integers comparing we obtain a + b = 4 + 3 = 7 which is our answer.

1^2= A(1)^3 + B(1)^2 + C(1) + D 1^2+3^2= A(2)^3 + B(2)^2 + C(2) + D 1^2+3^2+5^2= A(3)^3 + B(3)^2 + C(3) +D 1^2+3^2=5^2+7^2 = A(4)^3 + B(4)^2 + C(4) +D

and we'll get

1= A + B + C + D 10= 8A + 4B + 2C + D 35= 27A + 9B + 3C + D 84= 64A + 16B + 4C + D

By applying systems of equation

7A + 3B + C = 9 19A + 5B + C = 25 37A + 7B + C = 49

Continue until A is obtained

12A+2B=16 \rightarrow 6A=8 18A+2B=24 A= \frac {8}{6} = \frac {4}{3}

thus 4+3=7

Since the sequence can be expressed as a polynomial, we can use reduced row echelon form to figure out the coefficient of the first term, A.

We have to figure out the first few terms to do so, however. With that:

$a_1 = 1^2 = 1$

$a_2= 1^2 + 3^2 = 10$

$a_3 = 1^2 + 3^2 + 5^2 = 35$

$a_4 = 1^2 + 3^2 + 5^2 + 7^2 = 84$

We can then put this is a 4 x 5 matrix, evaluating n as the term in question. For example, we would plug in one for n in $An^3 + Bn^2 + Cn + D$ to find that the coefficients are $(1,1,1,1,1)$ and place that in the matrix.

This matrix would look like this:

{{1, 1, 1, 1, 1}, {8, 4, 2, 1, 10}, {27, 9, 3, 1, 35}, {64, 16, 4, 1, 84}}

Reduce the matrix using reduced row echelon form, and you recieve a matrix looking like this:

{{1, 0, 0, 0, 4/3}, {0, 1, 0, 0, 0}, {0, 0, 1, 0, -(1/3)}, {0, 0, 0,1, 0}}

A is the first number in the last column, and so we would add $4+3$ to find that the answer is 7.

Here is the notation. (x) ---- It is the equation obtained by substituting n=x Now u do (2)-(1) and get an equation.Let it be (A).

Now (3)-(2) and let it be (B). Now do (A)-(B). You will now get 12A+2B=16.

Then you go (4)-(3)-((3)-(1)) You will have 30A+4B=20. Thats it! Solve the both equations in A and B. A is found out to be 4/3. As the question says a=4 and b=3 and a+b=7

The sum of squares of first $n$ odd numbers is :

$1^2 + 3^2 + 5^2 + .................. + (2n-1)^2$

This is nothing but $\displaystyle \sum_{i=1}^n(2i-1)^2$

Now, expanding gives

$\displaystyle \sum_{i=1}^n(2i-1)^2 = \displaystyle \sum_{i=1}^n(4i^{2}- 4i + 1) = \displaystyle \sum_{i=1}^n4i^2 - \displaystyle \sum_{i=1}^n4i + \displaystyle \sum_{i=1}^n1 = 4\displaystyle \sum_{i=1}^ni^2-4\displaystyle \sum_{i=1}^ni + n.$

We know that $\displaystyle \sum_{i=1}^ni^2 = \dfrac{n(n+1)(2n+1)}{6}$ and $\displaystyle \sum_{i=1}^ni = \dfrac{n(n+1)}{2}.$

Now, substituting them, we get,

$4\displaystyle \sum_{i=1}^ni^2-4\displaystyle \sum_{i=1}^ni + n = \dfrac{4n(n+1)(2n+1)}{6} -\dfrac{4n(n+1)}{2}+n = \dfrac{2(2n^3+3n^2+n)}{3} - 2n^2 - 2n + n.$

Now, simplify it. We get,

$\dfrac{2(2n^3+3n^2+n)}{3} - 2n^2 - 2n + n = \dfrac{4n^3}{3}-\dfrac{n}{3}.$

Now, the coefficient of $n^3$ is $\dfrac{4}{3}.$

Therefore $\dfrac{a}{b}= \dfrac{4}{3} \Rightarrow a=4 , b=3 \Rightarrow a+b = 3+4 = \boxed{\boxed{7}}$

Finding the sum of odd squares is equivalent to finding the sum of all squares from 1 to $2n$ , and then subtracting off the even squares. We have

$\sum_{i=1}^{2n} i^2 = \frac{ 2n(2n+1)(4n+1) } { 6},$ $\sum_{i=1}^n (2i)^2 = 4 \sum_{i=1}^n i^2 = \frac{ 4n (n+1) (2n+1)} { 6}.$

Hence, $\sum_{i=1}^n (2i-1)^2 = \frac{ 2n(2n+1)(4n+1) } { 6} - \frac{ 4n (n+1) (2n+1)} { 6} .$

Thus, the coefficient of the cubic term is $\frac{ 2 \times 2 \times 4 } { 6} - \frac { 4 \times 1 \times 2 } { 6} = \frac{ 8}{6} = \frac{4}{3}$ . Thus, $a + b = 7$ .

Note: The sum of odd squares is equal to $\frac{ (2n-1)(2n)(2n+1) } { 6}$ .