Question

Show `1.2.3 + 2.3.4+ 3.4.5 + ... + n(n+1)(n+2) = 1/4n(n+1)(n+2)(n+3)`?

Answer 1

The sum is:

`S_n = 1.2.3 + 2.3.4+ 3.4.5 + ... + n(n+1)(n+2)`
`" " = sum_(r=1)^n r(r+1)(r+2)`
`" " = sum_(r=1)^n r(r^2+3r+2)`
`" " = sum_(r=1)^n (r^3+3r^2+2r)`
`" " = sum_(r=1)^n r^3 + 3sum_(r=1)^nr^2+2sum_(r=1)^nr`

We now use some standard formula for summation:

`sum_(r=1)^n r \ = 1/2n(n+1)`
`sum_(r=1)^n r^2 = 1/6n(n+1)(2n+1)`
`sum_(r=1)^n r^3 = 1/4n^2(n+1)^2`

Which gives us:

`S_n = 1/4n^2(n+1)^2 + 3*1/6n(n+1)(2n+1) + 2*1/2n(n+1)`
`" " = 1/4n^2(n+1)^2 + 1/2n(n+1)(2n+1) + n(n+1)`
`" " = 1/4n(n+1){n(n+1) +2(2n+1) + 4}`>
`" " = 1/4n(n+1)(n^2+n +4n+2 + 4)`
`" " = 1/4n(n+1)(n^2+5n+6)`
`" " = 1/4n(n+1)(n+2)(n+3) \ \` QED

Answer 2

We can also prove the given result using Mathematical Induction.

Let:

`S_n = 1.2.3 + 2.3.4+ 3.4.5 + ... + n(n+1)(n+2)`
`" " = sum_(r=1)^n r(r+1)(r+2)`

We want to prove that:

`S_n = 1/4n(n+1)(n+2)(n+3) \ \ \ AA n in NN`

Let use consider the case `n=1`. When `n=1` the given result gives:

`LHS = S_1`
`" " = 1.2.3`
`" " = 6`

And:

`RHS = 1/4(1)(1+1)(1+2)(1+3)`
`" " = (1.2.3.4)/4`
`" " = 24/4`
`" " = 6`

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

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

`S_k = sum_(r=1)^k r(r+1)(r+2) = 1/4k(k+1)(k+2)(k+3)`

Adding the next term to the series gives us:

`S_(k+1) = S_k + (k+1)(k+2)(k+3)`
`" " = 1/4k(k+1)(k+2)(k+3) + (k+1)(k+2)(k+3)` (by assumption)
`" " = (k(k+1)(k+2)(k+3))/4 + (4(k+1)(k+2)(k+3))/4`
`" " = 1/4(k+1)(k+2)(k+3)(k+4)`
`" " = 1/4 \ (k+1) \ {(k+1)+1} \ {(k+1)+2} \ {(k+1)+3}`

Which is the given result with `n=k+1`

So, we have shown that if the given result is true for `n=k`, then it is also true for `n=k+1`. But we initially showed that the given result was true for `n=1` and so it must also be true for `n=2, n=3, n=4, ...` and so on.

Hence, by the process of mathematical induction the given result is true for `n in NN` QED