Question

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

Answer 1

`S_n=47621`

Explanation:

We know that, `n^(th) term` of `"an "color(blue)"Arithmetic Progression "` is

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

`where,a=1^(st) term,and d="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=>4n-3=81=>4n=84=>n=21`

Using `(1)` we get

`S_n=sum_(r=1)^n T_r,where,T_r=(4r-3)^2 and n=21`

`=>S_n=sum_(r=1)^21 (4r-3)^2=sum_(r=1)^21(16r^2-24r+9)`

`=16sum_(r=1)^21 r^2-24sum_(r=1)^21 r+9sum_(r=1)^21 1`

Using `(2),(3),and(4)`

`S_n=16[n/6(n+1)(2n+1)]_(n=21) -24[n/2(n+1)]_21 +9xx21`

`=16*21/6(21+1)(2*21+1)-24*21/2(21+1)+189`

`=8xx7xx22xx43-12xx21xx22+189`

`=52976-5544+189`

`S_n=47621`

Answer 2

`47621`

Explanation:

Given:

`1^2+5^2+9^2+...+81^2`

Note this has `(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:

`color(blue)(1), 26, 107, 276`

The differences between consecutive terms are:

`color(blue)(25), 81, 169`

The differences of those differences are:

`color(blue)(56), 88`

The difference of those differences is:

`color(blue)(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)`

`color(white)(s_n) = 1+25n-25+28n^2-84n+56+16/3n^3-32n^2+176/3n-32`

`color(white)(s_n) = 1/3n(16n^2-12n-1)`

Then:

`s_21 = 1/3(color(blue)(21))(16(color(blue)(21))^2-12(color(blue)(21))-1)`

`color(white)(s_21) = 7(7056-252-1)`

`color(white)(s_21) = 7 * 6803`

`color(white)(s_21) = 47621`