Chapter 12.3, Problem 62SR

To determine

To prove: That the numbers of bricks in the top $n$ rows is $n^2+5n$ .

Explanation of Solution

Given information:

Each row has two more bricks than the row above it.

There are $6$ bricks in the top row.

Given

There are $6$ bricks in the top row.

It means $6$ can be written as

  $6=1^2+5\left(1\right)$

So, the formula $n^2+5n$ is True for $n=1$ .

Assume that $n^2+5n$ is true for $n=k$

There are $k^2+5k$ bricks in the top $k$ rows for some positive integer $k$ .

Since,

Each row has $2$ more bricks than the one above, the numbers of bricks in the rows from an arithmetic sequence.

The number of bricks in the ${\left(k+1\right)}^{th}$ row is $6+\left[\left(k+1\right)-1\right]\left(2\right)$

  $\Rightarrow$ The number of bricks in the ${\left(k+1\right)}^{th}$ row is $2k+6$ .

Therefore,

Number of bricks in the top $k+1$ rows is $k^2+5k+\left(2k+6\right)$

  $\Rightarrow$ Number of bricks in the top $k+1$ rows is $k^2+7k+6$

When we use given formula

  $n^2+5n$

On substituting $n=k+1$

We get

Number of bricks in top $k+1$ rows is ${\left(k+1\right)}^2+5\left(k+1\right)$

  $\Rightarrow$ Number of bricks in top $k+1$ rows is $k^2+2k+1+5k+5$

  $\Rightarrow$ Number of bricks in top $k+1$ rows is $k^2+7k+6$

Therefore,

We got same when substituting in given formula and through solving generally.

So, given

Number of bricks in top $n$ rows is $n^2+5n$ .