Binomial Theorem 1

Algebra Level 3

$\dbinom{n}{1}x(1-x)^{n-1}+2\dbinom{n}{2}x^2(1-x)^{n-2}+3\dbinom{n}{3}x^3(1-x)^{n-3}+ \ldots + n\dbinom{n}{n}x^n=Ax$

Find $A$ .

n

n+1

n-2

n-1

1 solution

Adarsh Kumar
Feb 9, 2016

$\text{Step 1:Finding the general term:}\\ \sum_{r=1}^{n} r\times \dbinom{n}{r}\times x^r\times (1-x)^{n-r}\\ \text{Step 2:Converting}\dbinom{n}{r} \text{to} \dfrac{n}{r}\times \dbinom{n-1}{r-1}\\ =\sum_{r=1}^{n}\left(r\times\dfrac{n}{r}\times \dbinom{n-1}{r-1} x^{r-1}\times (1-x)^{(n-1)-(r-1)}\right)\times x\\ =nx \times \sum_{r=1}^{n}[(x)+(1-x)]^{n-1}\\ \text{Step 3:Final answer:}nx \times 1^{n-1}\\ =nx,\text{hence} (...)=n$ .And done!

Moderator note:

Great. What's the motivation behind step 2?

Sorry for the late reply sir, @Calvin Lin I did that so as to remove the variable $r$ and replace it by a constant $n$ ,one easy way of doing that was to convert $\dbinom{n}{r}$ to a fraction containing $r$ in the denominator,hence this.:)

Adarsh Kumar - 5 years, 3 months ago

More generally, given a polynomial $f(r)$ , then $f(r) { n \choose r }$ can be written as $\sum g_i (n) { n \choose i }$ , where $g_i (n)$ are polynomials in $n$ . This allows us to simplify such summations.

E.g. Evaluate $\sum r^2 { n \choose r }$ .

Calvin Lin Staff - 5 years, 3 months ago

This particular problem can even be done by substituting x=1 since in LHS all terms are 0 except the last term which is n. Hence A=n.

Rishabh Jain - 5 years, 3 months ago

Binomial Theorem 1

1 solution

Moderator note:

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