Question

`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=`?

Answer 1

`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`