show by differentiation

**deej813** · September 1st 2009, 02:44 AM

if (1 + x)^n = (n/0) + (n/1)x + (n/2)x^2 + ........... + (n/n)x^n
show by differentiation that

2(2/n) + (2x3)(n/3) + (3x4)(n/4) + ......... + n(n-1)(n/n) = n(n-1)2^n-2

**HallsofIvy** · September 1st 2009, 03:48 AM

Originally Posted by deej813

if (1 + x)^n = (n/0) + (n/1)x + (n/2)x^2 + ........... + (n/n)x^n
show by differentiation that

2(2/n) + (2x3)(n/3) + (3x4)(n/4) + ......... + n(n-1)(n/n) = n(n-1)2^n-2

You mean $(1+ x)^n= \left(\begin{array}{c}n \\ 0\end{array}\right)+ \left(\begin{array}{c}n \\ 1\end{array}\right)x+ \left(\begin{array}{c}n \\ 2\end{array}\right)x^2+ \cdot\cdot\cdot +\left(\begin{array}{c}n \\ n\end{array}\right)x^n$
(what you wrote looks too much like fractions!)

Have you tried this at all? The " $2^{n-2}$ " on the right makes me think of differentiating twice (so that the power is reduced to n-2) and then taking x= 1 (so that x+1= 1+ 1= 2). What do you get if you do that on both sides?

**deej813** · September 1st 2009, 04:33 AM

yes thats what i mean
thanks
i don kow how to write them like that
um no i still dont get it

**Defunkt** · September 1st 2009, 05:35 AM

Originally Posted by deej813

if (1 + x)^n = (n/0) + (n/1)x + (n/2)x^2 + ........... + (n/n)x^n
show by differentiation that

2(2/n) + (2x3)(n/3) + (3x4)(n/4) + ......... + n(n-1)(n/n) = n(n-1)2^n-2

The equation we have is:

$(1+x)^n = \sum_{k=0}^n \binom{n}{k}x^k 1^{n-k} = \binom{n}{0}x^0 + \binom{n}{1}x + ... +\binom{n}{n-1}x^{n-1}+ \binom{n}{n}x^n$

Differentiating both sides once, we get:

$n(x+1)^{n-1} = \binom{n}{1} + 2\binom{n}{2}x + ... + (n-1)\binom{n}{n-1}x^{n-2} + n\binom{n}{n}x^{n-1}$

Differentiating again, we get:

$n(n-1)(x+1)^{n-2}= 2\binom{n}{2} + 2 \cdot 3\binom{n}{3}x + ...$ $+ (n-1)(n-2)\binom{n}{n-1}x^{n-3} + n(n-1)\binom{n}{n}x^{n-2}$

If we take x = 1, we get:

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

**HallsofIvy** · September 1st 2009, 06:10 AM

Originally Posted by deej813

yes thats what i mean
thanks
i don kow how to write them like that
um no i still dont get it

Do you not know that the derivative of $x^n$ is $n x^{n-1}$ ? (For n not equal to -1.) That's usually the first derivative you learn!

**deej813** · September 1st 2009, 12:41 PM

yeah i do get that
um i get most of it
just after the first differentiation where does the +(n-1)(n/n-1)x^n-2 come from
sorry i wrote it like fractions again

**Defunkt** · September 1st 2009, 12:51 PM

Originally Posted by deej813

yeah i do get that
um i get most of it
just after the first differentiation where does the +(n-1)(n/n-1)x^n-2 come from
sorry i wrote it like fractions again

What's the derivative of $\binom{n}{k}x^t$ , with respect to $x$ ?

**deej813** · September 1st 2009, 01:31 PM

isnt it
t(n/k)x^t-1

**Defunkt** · September 1st 2009, 01:36 PM

Originally Posted by deej813

isnt it
t(n/k)x^t-1

Correct. The term you asked about - I just did not specifically write it in the initial equation. Of course, this does not mean it was not there!

I fixed it now so it is clearer. Post here if you still need clarification!

**deej813** · September 1st 2009, 10:15 PM

ok great
i get it now

thanks heaps

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