PROVE BY INDUCTION

**poppy12345** · December 10th 2008, 04:48 AM

I have a question and am unsure how to start it:

Prove by induction that
(sum from k=1 to n)of
k(k+1)...(k+a) = 1/(a+2)*n(n+1)(n+2)....(n+a+1)

**Grandad** · December 10th 2008, 07:19 AM

Hello -

Originally Posted by poppy12345

I have a question and am unsure how to start it:

Prove by induction that
(sum from k=1 to n)of
k(k+1)...(k+a) = 1/(a+2)*n(n+1)(n+2)....(n+a+1)

Tricky to know where to start, isn't it? Is it $n$ that we vary, or $a$ ?

Well, start in the usual way, by writing a proposition about $n$ , and let $a$ take care of itself. So:

Let $S_n = \sum_{k=1}^n k(k+1)...(k+a)$

Then the propositional function $P(n)$ is defined as:

$P(n)\equiv S_n = \frac{1}{a+2}n(n+1)(n+2)...(n+a+1)$

Then write down the sum $S_{n+1}$ as $S_n +$ the term you get when $k=(n+1)$ .

You'll find you can then take out lots of common factors, and eventually express this as $\sum_{k=1}^{n+1} k(k+1)...(k+a)$ , thus showing that $P(n)\implies P(n+1)$ .

Finally, you'll need to prove that $P(1)$ is true for any $a$ .

I hope I've given you enough to go on. Let me know if you need more help.

Grandad

**james_bond** · December 10th 2008, 07:43 AM

Suppose $\sum_{k=1}^n\prod_{i=0}^a(k+i)=\frac 1{a+2}\prod_{j=0}^{a+1}(n+j)$ . So we need to show that $\sum_{k=1}^{n+1}\prod_{i=0}^a(k+i)=\frac 1{a+2}\prod_{j=0}^{a+1}(n+1+j)$ .
$\sum_{k=1}^{n+1}\prod_{i=0}^a(k+i)=\sum_{k=1}^n\pr od_{i=0}^a(k+i)+\prod_{i=0}^a(n+1+i)=\frac 1{a+2}\prod_{j=0}^{a+1}(n+j)+\prod_{i=0}^a(n+1+i)$ by the assumption. So this should be equal to $\frac 1{a+2}\prod_{j=0}^{a+1}(n+1+j)$ : $\frac 1{a+2}\prod_{j=0}^{a+1}(n+j)+\prod_{i=0}^a(n+1+i)= \frac 1{a+2}\prod_{j=0}^{a+1}(n+1+j)\leftrightarrow$ $\frac 1{a+2}\prod_{j=0}^{a+1}(n+j)=\frac 1{a+2}\prod_{j=0}^{a+1}(n+1+j)-\prod_{i=0}^a(n+1+i)=$ $\left[\prod_{i=0}^a\left(n+1+i\right)\right]\cdot\left(\frac 1{a+2}\cdot \left(n+1+(a+1)\right)-1\right)\leftrightarrow$ $\prod_{j=0}^{a+1}(n+j)=\left[\prod_{i=0}^a\left(n+1+i\right)\right]\cdot n$ $\leftrightarrow n\cdot (n+1)\cdot\ldots\cdot (n+a+1)=\left[(n+1)\cdot (n+2)\cdot\ldots\cdot (n+a+1)\right]\cdot n$ QED

**mitch_nufc** · December 10th 2008, 07:51 AM

DR Duncans class by any chance?

**poppy12345** · December 11th 2008, 01:19 AM

yea

**mitch_nufc** · December 11th 2008, 06:19 AM

are you still stuck on that or anything? i handed in today, pretty confident everythings right

**poppy12345** · December 11th 2008, 07:17 AM

well i think ive managed but going tomorro. never very confident but neva mind!

**mitch_nufc** · December 11th 2008, 07:43 AM

im always happy to help if ya stuck... just shout, whats ur name?

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