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Find the generating function for the sequence $\left \{ a_n \right \} where$

$a_n = \Large \binom{10}{n+1}$ for n = 0,1,2,….

Sol.

$\Large \binom{10}{1} + \binom{10}{2}x + \binom{10}{3}x^2 + \binom{10}{4} x^3 + .... + \binom{10}{10}x^{9}$

multiplying and dividing above equation by x,

$\Large \frac{1}{x} (\binom{10}{1}x + \binom{10}{2}x^2 + \binom{10}{3}x^3 + \binom{10}{4} x^4 + .... + \binom{10}{10}x^{10})$

adding and subtracting $\large \frac{1}{x}$,

$\Large \frac{1}{x} – \frac{1}{x} + \frac{1}{x} (\binom{10}{1}x + \binom{10}{2}x^2 + \binom{10}{3}x^3 + \binom{10}{4} x^4 + .... + \binom{10}{10}x^{10})$

$\Large – \frac{1}{x} + \frac{1}{x} (1 + \binom{10}{1}x + \binom{10}{2}x^2 + \binom{10}{3}x^3 + \binom{10}{4} x^4 + .... + \binom{10}{10}x^{10})$

Using binomial theorem,

$\Large – \frac{1}{x} + \frac{1}{x} ( 1+x)^{10}$

$\Large \color{red}{ \frac{( 1+x )^{10} – 1}{x} }$

edited | 45 views
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yes, right
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thanks for confirming :)

Yes,you are absolutely correct,I just want to differ on the part of getting to the ans from the second step which I found a little easier.

multiplying and dividing above equation by x,we get:-

$\frac{1}{x} (\begin{pmatrix} 10\\ 1 \end{pmatrix}x^{1}1^{10-1} + \begin{pmatrix} 10\\ 2 \end{pmatrix}x^{2}1^{10-2} + \begin{pmatrix} 10\\ 3 \end{pmatrix}x^{3}1^{10-3} ................... \begin{pmatrix} 10\\ n \end{pmatrix}x^{n}1^{10-n})$

Now,let's add and subtract 1 inside the braces as follows:-

$\frac{1}{x} (1+\begin{pmatrix} 10\\ 1 \end{pmatrix}x^{1}1^{10-1} + \begin{pmatrix} 10\\ 2 \end{pmatrix}x^{2}1^{10-2} + \begin{pmatrix} 10\\ 3 \end{pmatrix}x^{3}1^{10-3} ................... \begin{pmatrix} 10\\ n \end{pmatrix}x^{n}1^{10-n}-1)$

Now,a binomial pattern can be observed to be emerging from the above manipulation.Let's make it a full formed binomial expression by replacing first 1 as below:

$\frac{1}{x} (\begin{pmatrix} 10\\ 0 \end{pmatrix}x^{0}1^{10-0}+\begin{pmatrix} 10\\ 1 \end{pmatrix}x^{1}1^{10-1} + \begin{pmatrix} 10\\ 2 \end{pmatrix}x^{2}1^{10-2} + \begin{pmatrix} 10\\ 3 \end{pmatrix}x^{3}1^{10-3} ....... \begin{pmatrix} 10\\ n \end{pmatrix}x^{n}1^{10-n}-1)$

The above binomial expansion can be rewritten as:

$\frac{1}{x} ((1+x)^{10}-1 )$

The second step is exactly same as yours,just doing off with the fractions and keeping all the 1's in the expression just made it a little easier to follow for me,I hope it helps others too. :)
by Junior (879 points)

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