Question

Refer to the attachment and answer these questions.

( 1 ). Find the first counting number 'x' in group n.
( 2 ). Find the sum S of all counting numbers in group n.
Please also show your work too so others and I can understand. Thanks in advance!

Answer 1

`\displaystyle \text{(1)}\\x=1+\sum_(k=1)^(n-1)2k\\\\\text{(2)}\\\displaystyle (\displaystyle \sum_(k=1)^(n)2k\left(1+\displaystyle \sum_(k=1)^(n)2k \right))/(2)-(\displaystyle \sum_(k=1)^(n-1)2k\left(1+\displaystyle \sum_(k=1)^(n-1)2k \right))/(2)`

Explanation:

Notice that Group 1 starts with 1 and we keep counting up by 1.

Part 1:

The total number of numbers from Group 1 to Group
`n` can be found by:

`\displaystyle \sum_(k=1)^(n)2k`

To find the first number in a certain group
`g`, we'll find the total number of numbers from Group 1 to Group
`g-1`, then add 1.

Here's why.

Let's say we want to find the first number of Group 3, which we can clearly see is 7. Take advantage of the fact that the sequence of numbers counts up by 1, starting from 1. We will use the formula above to to find the total number of numbers from Group 1 to Group 2, which is 6, then add 1 to achieve the number of 7.

Therefore, the first number in group
`n` must be:

`\boxed{\displaystyle 1+\sum_(k=1)^(n-1)2k}`

Part 2:

Recall that the sum of the numbers from
`1` to positive integer
`n` inclusive is given by
`\displaystyle(n(n+1))/(2)`.

Therefore, if we find the total number of numbers from Group 1 to Group
`n`, we can simply substitute that value into the formula above to find the total sum of the numbers from Group 1 to Group
`n`.

To find the total sum of the numbers exclusively in Group
`n` simply subtract the sum of numbers from Group 1 to Group
`n-1`.

Short example on how that works:

The sum of all numbers in Group 3 is equal to the sum of the numbers from Group 1 to Group 3 minus the sum of the numbers from Group 1 to Group 2.

We've already found the formula for the total number of numbers from Group 1 to Group
`n` in the previous part, so convert this concept into summation notation like such:

`\boxed{\displaystyle (\displaystyle \sum_(k=1)^(n)2k\left(1+\displaystyle \sum_(k=1)^(n)2k \right))/(2)-(\displaystyle \sum_(k=1)^(n-1)2k\left(1+\displaystyle \sum_(k=1)^(n-1)2k \right))/(2)}`