#a_1=1/1^2# #a_2=1/2^2+2/2^2# #a_3=1/3^2+2/3^2+3/3^2# #a_4=1/4^2+2/4^2+3/4^2+4/4^2# #a_n=#?

1 Answer
Mar 19, 2018

#a_color(blue)(n)=1/color(blue)(n)^2+2/color(blue)(n)^2+3/color(blue)(n)^2+4/color(blue)(n)^2 +5/color(blue)(n)^2 + ..........+ n/color(blue)(n)^2#

Explanation:

This question is more about observation than doing maths,

Look for the pattern that is given with the rows of numbers and then do the same with #n#
#a_color(blue)(1)=1/color(blue)(1)^2#
#a_color(blue)(2)=1/color(blue)(2)^2+2/color(blue)(2)^2#
#a_color(blue)(3)=1/color(blue)(3)^2+2/color(blue)(3)^2+3/color(blue)(3)^2#
#a_color(blue)(4)=1/color(blue)(4)^2+2/color(blue)(4)^2+3/color(blue)(4)^2+4/color(blue)(4)^2#

What do we see?

  • each term consists of a fraction
  • the numerators increase by #1# each time
  • each value in the denominator is squared
  • the denominators change, but the number is the same as the row and the same as the subscript of the #a# on the left hand side.
  • the number of terms is the same as the subscript.
  • in the last term the numerator is the same as the subscript

Continuing the pattern would give the #5th# row as:

#a_color(blue)(5)=1/color(blue)(5)^2+2/color(blue)(5)^2+3/color(blue)(5)^2+4/color(blue)(5)^2 +5/color(blue)(5)^2#

For the #nth# row, we do not know the value of #n# so we do not know how far to continue, but we can show the same pattern as is described above.

#a_color(blue)(n)=1/color(blue)(n)^2+2/color(blue)(n)^2+3/color(blue)(n)^2+4/color(blue)(n)^2 +5/color(blue)(n)^2 + ..........+ n/color(blue)(n)^2#