# Find the sum of 1^2 + 5^2 + 9^2 + … 81^2?

Apr 19, 2018

${S}_{n} = 47621$

#### Explanation:

We know that, ${n}^{t h} t e r m$ of $\text{an "color(blue)"Arithmetic Progression }$ is

color(blue)(t_n=a+(n-1)d,to(I^star)

$w h e r e , a = {1}^{s t} t e r m , \mathmr{and} d = \text{common difference} = {t}_{n} - {t}_{n - 1}$

Also, note that

color(red)((1)S_n=sum_(r=1)^n T_r

color(red)((2)sum_(r=1)^n 1=n

color(red)((3)sum_(r=1)^n r=n/2(n+1)

color(red)((4)sum_(r=1)^n r^2=n/6(n+1)(2n+1)

WE have,

S_n=1^2 + 5^2 + 9^2 + … 81^2

The Arithmetic sequence is:

1,5,9,...81.=>color(blue)(a=1,d=4,t_n=81,n= ?.apply (I^star)

:.color(green)(t_n)=1+(n-1)4=1+4n-4=color(green)(4n-3

But ,${t}_{n} = 81 \implies 4 n - 3 = 81 \implies 4 n = 84 \implies n = 21$

Using $\left(1\right)$ we get

${S}_{n} = {\sum}_{r = 1}^{n} {T}_{r} , w h e r e , {T}_{r} = {\left(4 r - 3\right)}^{2} \mathmr{and} n = 21$

$\implies {S}_{n} = {\sum}_{r = 1}^{21} {\left(4 r - 3\right)}^{2} = {\sum}_{r = 1}^{21} \left(16 {r}^{2} - 24 r + 9\right)$

$= 16 {\sum}_{r = 1}^{21} {r}^{2} - 24 {\sum}_{r = 1}^{21} r + 9 {\sum}_{r = 1}^{21} 1$

Using $\left(2\right) , \left(3\right) , \mathmr{and} \left(4\right)$

${S}_{n} = 16 {\left[\frac{n}{6} \left(n + 1\right) \left(2 n + 1\right)\right]}_{n = 21} - 24 {\left[\frac{n}{2} \left(n + 1\right)\right]}_{21} + 9 \times 21$

$= 16 \cdot \frac{21}{6} \left(21 + 1\right) \left(2 \cdot 21 + 1\right) - 24 \cdot \frac{21}{2} \left(21 + 1\right) + 189$

$= 8 \times 7 \times 22 \times 43 - 12 \times 21 \times 22 + 189$

$= 52976 - 5544 + 189$

${S}_{n} = 47621$

Apr 19, 2018

$47621$

#### Explanation:

Given:

${1}^{2} + {5}^{2} + {9}^{2} + \ldots + {81}^{2}$

Note this has $\frac{81 - 1}{4} + 1 = 21$ terms

Note also that the sum to $n$ terms will be given by a cubic formula in $n$.

Hence if we can find a cubic formula that matches the first four sums, then it will be correct for any number of terms.

The first four sums are:

$\textcolor{b l u e}{1} , 26 , 107 , 276$

The differences between consecutive terms are:

$\textcolor{b l u e}{25} , 81 , 169$

The differences of those differences are:

$\textcolor{b l u e}{56} , 88$

The difference of those differences is:

$\textcolor{b l u e}{32}$

We can then use the initial term of each of these sequences as coefficients to provide a formula:

s_n = color(blue)(1)/(0!)+color(blue)(25)/(1!)(n-1) + color(blue)(56)/(2!)(n-1)(n-2)+color(blue)(32)/(3!)(n-1)(n-2)(n-3)

$\textcolor{w h i t e}{{s}_{n}} = 1 + 25 n - 25 + 28 {n}^{2} - 84 n + 56 + \frac{16}{3} {n}^{3} - 32 {n}^{2} + \frac{176}{3} n - 32$

$\textcolor{w h i t e}{{s}_{n}} = \frac{1}{3} n \left(16 {n}^{2} - 12 n - 1\right)$

Then:

${s}_{21} = \frac{1}{3} \left(\textcolor{b l u e}{21}\right) \left(16 {\left(\textcolor{b l u e}{21}\right)}^{2} - 12 \left(\textcolor{b l u e}{21}\right) - 1\right)$

$\textcolor{w h i t e}{{s}_{21}} = 7 \left(7056 - 252 - 1\right)$

$\textcolor{w h i t e}{{s}_{21}} = 7 \cdot 6803$

$\textcolor{w h i t e}{{s}_{21}} = 47621$