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

2 Answers
Apr 19, 2018

#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#

Apr 19, 2018

#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#