Question

The first four figures of a pattern are shown below. How many squares would be needed to build the nth figure?

Answer 1

`2n^2-2n+1`

Explanation:

let ‘n’ be the number of the figure.

For n = 2 :

The number of the squares of the second figure :

= (the number of the squares of the middle/horizontal row) + (the number of the remaining squares)

= (2×2−1) + 2×(1+2×0)

= (3) + 2×(1)

= 3 + 2

= 5

For n = 3 :

The number of the squares of the third figure :

= (the number of the squares of the middle/horizontal row) + (the number of the remaining squares)

= (2×3−1) + 2×[(1+2×0) + (1+2×1)]

= (5) + 2×[(1) + (3)]

= 5 + 2×[4]

= 5 + 8

= 13

For n = 4 :

The number of the squares of the fourth figure :

= (the number of the squares of the middle/horizontal row) + (the number of the remaining squares)

= (2×4−1) + 2×[(1+2×0) + (1+2×1)+ (1+2×2)]

= (7) + 2×[(1) + (3) + (5)]

= 7 + 2×[9]

= 7 + 18

= 25

For n (n ≥ 2) :

The number of the squares of the nth figure :

= (the number of the squares of the middle/horizontal row) + (the number of the remaining squares)

`= (2n-1) +2* \left( \sum^(n-2)_(k=0) \left( 1+2k\right) \right)`

`= (2n-1) +2* \left( \sum^(n-2)_(k=0) \left( 1\right) \right)+4* \left( \sum^(n-2)_(k=0) \left( k\right) \right)`

`= (2n-1) +2(n-1)+4* [(n-1)/(2)(0+(n-2))]`

`= (2n-1) +(2n-2)+4* [(n-1)/(2)(n-2)]`

`= 4n-3+2* (n-1)(n-2)`

`= 4n-3+2* (n^2-3n+2)`

`= 4n-3+2n^2-6n+4`

`= 2n^2-2n+1`