Question

Simplifying `S_(k+1)` completely. Thanks?!!

Answer 1

`S_k=[k(k+1)(k+2)]/3`

`S_(k+1)=[(k+1)(k+2)(k+3)]/3`

Explanation:

Can't we just substitute `x=k+1` into the formula, or am I missing something here?

The sequence is:

`S_n=1*2+2*3+3*4+...+n(n+1)=[n(n+1)(n+2)]/3`

So, if we want to calculate `S_k`, we just put `n=k`, and get

`S_k=1*2+2*3+3*4+...+k(k+1)=[k(k+1)(k+2)]/3`

In the case of `S_(k+1)`, I think we can just substitute `n=k+1`, and we are going to have

`S_(k+1)=1*2+2*3+3*4+...+(k+1)(k+2)=[(k+1)(k+2)(k+3)]/3`

If we want to expand this, it becomes

`[(k+1)(k+2)(k+3)]/3`

`=[(k^2+3k+2)(k+3)]/3`

`=(k^3+3k^2+3k^2+9k+2k+6)/3`

`=(k^3+6k^2+11k+6)/3`

`=k^3/3+(6k^2)/3+(11k)/3+6/3`

`=k^3/3+2k^2+(11k)/3+2`

Answer 2

`S_(k+1)=((k+1)(k+2)(k+3))/3`

Explanation:

`S_n:1.2+2.3+3.4+...+n(n+1)=(n(n+1)(n+2))/3`

Let the statement be true for n=k,
`S_k:1.2+2.3+3.4+...+k(k+1)=(k(k+1)(k+2))/3`

Let us verify for
n=k+1, then
`S_n=S_(k+1)`
`n+1=k+2`
`n+2=k+3`
`"with the immediate term being" (k+1)(k+2)`

`(n(n+1)(n+2))/3=((k+1)(k+2)(k+3))/3`

Thus,

`S_(k+1):1.2+2.3+3.4+...+k(k+1)+(k+1)(k+2)`

`S_(k+1):S_k+(k(k+1)(k+2))/3`

`=(k(k+1)(k+2))/3+(k+1)(k+2)`
`=1/3(k(k+1)(k+2)+3(k+1)(k+2))`

`=1/3((k+1)(k+2)(k+3))=((k+1)(k+2)(k+3))/3`

Verified.

Thus
`S_(k+1)=((k+1)(k+2)(k+3))/3`