summation and mathematical induction

**wintersoltice** · July 15th 2010, 05:15 AM

1.Prove by induction that $\sum_{r=1}^{n}(\frac{1}{r(r+1)(r+2)})$ $=$ $\frac{n(n+3)}{4(n+1)(n+2)}$ .

Show that $\frac{n(n+3)}{4(n+1)(n+2)}$ is LESSER than $\frac{1}{4}$ for all positive integer values of n.

Deduce from these results that $\sum_{r=1}^{n}(\frac{1}{(r+1)^3})$ is LESSER than $\frac{1}{4}$ .

i can do the first part of the question. i need help for the second and third parts in getting started.

**Archie Meade** · July 15th 2010, 07:23 AM

Originally Posted by wintersoltice

1.Prove by induction that $\sum_{r=1}^{n}(\frac{1}{r(r+1)(r+2)})$ $=$ $\frac{n(n+3)}{4(n+1)(n+2)}$ .

Show that $\frac{n(n+3)}{4(n+1)(n+2)}$ is LESSER than $\frac{1}{4}$ for all positive integer values of n.

Deduce from these results that $\sum_{r=1}^{n}(\frac{1}{(r+1)^3})$ is LESSER than $\frac{1}{4}$ .

i can do the first part of the question. i need help for the second and third parts in getting started.

For the second part

$\frac{1}{4}\left(\frac{n(n+3)}{(n+1)(n+2)}\right)< \frac{1}{4}$ if n(n+3)<(n+1)(n+2)

$n^2+3n<n^2+3n+2\ ?$ yes, it is

**Archie Meade** · July 15th 2010, 08:32 AM

Originally Posted by wintersoltice

1.Prove by induction that $\sum_{r=1}^{n}(\frac{1}{r(r+1)(r+2)})$ $=$ $\frac{n(n+3)}{4(n+1)(n+2)}$ .

Show that $\frac{n(n+3)}{4(n+1)(n+2)}$ is LESSER than $\frac{1}{4}$ for all positive integer values of n.

Deduce from these results that $\sum_{r=1}^{n}(\frac{1}{(r+1)^3})$ is LESSER than $\frac{1}{4}$ .

i can do the first part of the question. i need help for the second and third parts in getting started.

The third part involves showing that

$\frac{1}{(r+1)^3}<\frac{1}{r(r+1)(r+2)}$

$\frac{1}{(r+1)^2}<\frac{1}{r(r+2)}\ ?$

$\frac{1}{r^2+2r+1}<\frac{1}{r^2+2r}\ ?$

$r^2+2r+1>r^2+2r\ ?$

The induction proof appears more challenging than parts 2 and 3.
Here's how I would give the Induction Proof....

P(k)

$\sum_{r=1}^{k}\frac{1}{r(r+1)(r+2)}=\frac{k(k+3)}{ 4(k+1)(k+2)}\ ?$

P(k+1)

$\sum_{r=1}^{k+1}\frac{1}{r(r+1)(r+2)}=\frac{(k+1)( k+4)}{4(k+2)(k+3)}\ ?$

Try to prove that P(k) being true causes P(k+1) to also be true.

Proof

$\sum_{r=1}^{k+1}\frac{1}{r(r+1)(r+2)}=\sum_{r=1}^{ k}\frac{1}{r(r+1)(r+2)}+\frac{1}{(k+1)(k+2)(k+3)}$

$=\frac{k(k+3)}{4(k+1)(k+2)}+\frac{1}{(k+1)(k+2)(k+ 3)}$

if P(k) is true.

This gives

$\frac{k(k+3)^2}{4(k+1)(k+2)(k+3)}+\frac{4}{k+1}\le ft(\frac{1}{4(k+2)(k+3)}\right)$

$\frac{1}{4(k+2)(k+3)}\left(\frac{k(k+3)^2}{k+1}+\f rac{4}{k+1}\right)$

$=\frac{1}{4(k+2)(k+3)}\left(\frac{k\left(k^2+6k+9\ right)+4}{k+1}\right)$

$=\frac{1}{4(k+2)(k+3)}\left(\frac{k^3+6k^2+9k+4}{k +1}\right)$

$=\frac{1}{4(k+2)(k+3)}\frac{(k+1)\left(k^2+5k+4\ri ght)}{k+1}$

$=\frac{(k+4)(k+1)}{4(k+2)(k+3)}$

Then test your initial value to complete the proof.

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