Using Binomial Theorem in proving Combinatorial identities « RP

9 May

Using Binomial Theorem in proving Combinatorial identities

Posted May 9, 2010 by rosapaulina in Mathematics. Tagged: binomial, binomial theorem, combination, combinatorics, identity, Math, Mathematics, permutation, summation. 2 Comments

1) Prove:

$\binom{n}{0} + 8\binom{n}{1}+8^2\binom{n}{2}+8^3\binom{n}{3}+.....+8^n\binom{n}{n}=9^n$

Solution:

The above expression can be simplified as
$\sum_{r=0}^{n} 8^r \binom{n}{r}$

Applying binomial theorem,
$(x+y)^n=\sum_{r=0}^{n}\binom{n}{r}x^{n-r}y^r$

setting x=1 and y=8
$(1+8)^n=\sum_{r=0}^{n}\binom{n}{r} 1^{n-r}8^r$

Thus,
$\sum_{r=0}^{n} 8^r\binom{n}{r}=9^n$

2) Show that

$\binom{n}{0}+\frac{1}{2}\binom{n}{1}+\frac{1}{3}\binom{n}{2}+\frac{1}{4}\binom{n}{3}+.....+\frac{1}{n+1}\binom{n}{n}=\frac{1}{n+1}(2^{n+1}-1)$

Solution:

The simplified version of the above expression is
$\sum_{r=0}^{n}\frac{1}{r+1}\binom{n}{r}$

Using binomial theorem,

$(x+y)^n=\sum_{r=0}^{n}\binom{n}{r}x^{n-r}y^r$ Letting x=1
$(1+y)^n=\sum_{r=0}^{n}\binom{n}{r}y^r$

Integrating both sides,
$\frac{(1+y)^{n+1}}{n+1} + C = \sum_{r=0}^{n}\frac{1}{r+1}\binom{n}{r}y^{r+1}$

setting y=1
$\sum_{r=0}^{n}\frac{1}{r+1}\binom{n}{r}=\frac{2^{n+1}}{n+1}+C$

The last step is solving for C
If y=0, then
$\frac{1}{n+1}+C=0$

$C=-\frac{1}{n+1}$

Therefore,
$\sum_{r=0}^{n}\frac{1}{r+1}\binom{n}{r}=\frac{2^{n+1}}{n+1}-\frac{1}{n+1}$

$\sum_{r=0}^{n}\frac{1}{r+1}\binom{n}{r}=\frac{1}{n+1}(2^{n+1}-1)$

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2 responses to this post.

Posted by Michael Corpus on May 10, 2010 at 12:46 PM

hi RP!!

ang galing2 mo talaga!!

ganda ng blog mo.,

keep up the good work!!

=)

Reply
- Posted by rosapaulina on May 10, 2010 at 12:48 PM
  
  Uy Michael! salamat
  
  gawa ka din dito
  
  Reply

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