Question

How to write `S_1, S_2, S_3`, and show that each of these statements is true? Thanks!

Answer 1

Follow the instructions.

Explanation:

Write `S_1`

`S_1 :` `2` is a factor of `(1)^2+7(1)`

How do you prove that `2` is a factor of `8`?

You point out that `8 = 2xx4`.

Now do something similar for `S_2` and `S_3`.

Answer 2

Simply plug in 1, 2, and 3 into `n` and show for each case that 2 is a factor. See below for full explanation.

Explanation:

The problem is structured a bit odd, but it is pretty straightforward.

`S_1: 2 " is a factor of " 1^2 + 7*1`
`1^2 + 7*1 = 8`
`2` is a factor of `8` because `8` is divisible by `2`.

`S_2: 2 " is a factor of " 2^2 + 7*2`
`2^2 + 7*2 = 18`
`2` is a factor of `18` because `18` is divisible by `2`.

`S_3: 2 " is a factor of " 3^2 + 7*3`
`3^2 + 7*3 = 30`
`2` is a factor of `30` because `30` is divisible by `2`.

Unless I am missing some context, that should be all there is to it.

`square`

Answer 3

We have:

`S_n: 2 " is a factor of "n^2+7n`

So then statements `S_1, S_2 and S_3` would be:

`S_1: 2 " is a factor of "1^2+7=8`, which is true .

`S_2: 2 " is a factor of "2^2+14=18`, which is true .

`S_3: 2 " is a factor of "3^2+21=30`, which is true .

In general, we can prove the result is true using Induction:

Induction Proof - Hypothesis

We seek to prove that:

`S_n: 2 " is a factor of "n^2+7n` ..... [A]

So let us test this assertion using Mathematical Induction:

Induction Proof - Base case:

We will show that the given result, [A], holds for `n=1`

When `n=1` we have `S_1`, which we have already shown to be true/

So the given result is true when `n=1`.

Induction Proof - General Case

Now, Let us assume that the given result [A] is true when `n=m`, for some `m in NN, m gt 1`, in which case for this particular value of `m` we have:

`2 " is a factor of "m^2+7m`

In other words:

`m^2+7m = 2A` for som `A in NN` ..... [B]

Consider the expression:

`(m+1)^2+7(m+1) = (m^2+2m+1)+7(m+1)`
`" "= m^2+2m+1+7m+7`
`" "= (m^2+7m)+2m+8`
`" "= 2A+2(m+4)`
`" "= 2{Am+4}`

Which is the given result [A] with `n=m+1`

Induction Proof - Summary

So, we have shown that if the given result [A] is true for `n=m`, then it is also true for `n=m+1` (ie, `S_m => S_(m+1)`) where `m gt 1`. But we initially showed that the given result was true for `n=1` so it must also be true for `n=2, n=3, n=4, ...` and so on.

Induction Proof - Conclusion

Then, by the process of mathematical induction the given result [A] is true for `n in NN`

Hence we have:

`S_n: 2 " is a factor of "n^2+7n \ \ \` QED

Answer 4

Please see below.

Explanation:

`S_1-2" is a factor of "1^2+7*1`
`->` as `1^2+7*1=1+7=8`

`S_2-2" is a factor of "2^2+7*2`
`->` as `2^2+7*2=4+14=18`

`S_3-2" is a factor of "3^2+7*3`
`->` as `3^2+7*3=9+21=30`

In fact this is true for all `n` as

`n^2+7n=n(n+7)` is always divisible by `2` for following reason.

Now if `n` is odd, `n+7` is even and `n(n+7)` is even

and if `n` is even, `n+7` is odd and `n(n+7)` is even