Binomial theorem Question! (1 Viewer)

[Premus]

The coefficient of x^k in (1+x)^n , where n is a positive integer , is denoted by n C k same as (c k)

a) show that


c0 + 2 c1 + 3 c2 + .... + (n+1) cn = (n+2) 2 ^ (n-1)


b) find the sum

c0/ 1*2 - c1/2*3 + c2 / 3*4 - .... (-1)^n . cn / (n+1)(n+2)


Write your answer as a simple expression in terms of n.

btw - this question is Question 7 from the 2002 HSC paper

Thanks!!

 

[CM_Tutor]

(a) Putting x = 1 into the expansion of (1 + x)ⁿ gives: c0 + c1 + c2 + ... + cn = 2ⁿ _____ (1)

Differentiating the expansion of (1 + x)ⁿ, and then putting x = 1 gives:
c1 + 2 * c2 + 3 * c3 + ... + n * cn = n * 2^n-1 _____ (2)

(1) + (2) gives c0 + 2 * c1 + 3 * c2 + ... + (n + 1) * cn = (n + 2) * 2^n-1

(b) Integrating the expansion of (1 + x)ⁿ, and evaluating the constant by putting x = 0, gives:

c0 * x + c1 * x² / 2 + c2 * x³ / 3 + ... + cn * xⁿ⁺¹ / (n + 1) = [(1 + x)ⁿ⁺¹ - 1] / (n + 1)

Integrating this a second time, again evaluating the constant by putting x = 0.

The required sum can then be found by putting x = 1.

 

[Premus]

Thanks!! i get part a)

but for part b)....
i can integrate the expansion to get

c0 * x + c1 * x2 / 2 + c2 * x3 / 3 + ... + cn * xn+1 / (n + 1)


...but i dont know how u get

[(1 + x)n+1 - 1] / (n + 1)
or what you did after that

Thanks for ur help!

 

[CM_Tutor]

Originally posted by PremusDog
...but i dont know how u get

[(1 + x)n+1 - 1] / (n + 1)
or what you did after that

Integrate (1 + x)ⁿ. You get (1 + x)ⁿ⁺¹ / (n + 1) + C, for some constant C

Putting x = 0, the LHS becomes 0 + 0 + 0 + ... + 0 = 0

The RHS becomes (1 + 0)ⁿ⁺¹ / (n + 1) + C = 1 / (n + 1) + C.

It follows that C = -1 / (n + 1), and hence the RHS is
[(1 + x)ⁿ⁺¹ / (n + 1)] + [-1 / (n + 1)] = [(1 + x)ⁿ⁺¹ - 1] / (n + 1)

 

[Premus]

Thanks!!

 

[psych_girl]

I am bit confused about finding the greatest constant term.

 

[Estel]

A.) Give an example.

B.) how can you tutor 2U/3U?

 

[mojako]

well, suppose it turns out that T_(k+1) / T_k > 1 when k < 4.2
so, T_(k+1) is bigger than T_k up to k = 4
and the biggest term is T_(k+1) which is T_(4+1) = T_5

Or you can just remember to round the number up, or add 1 if its a whole number
4.2 becomes 5 and 4 becomes 5, and 4.9 also becomes 5.
EDIT: WRONG! if its a whole number then it stays the same.. if k<4 then the greatest term is T_4

If you use T_k / T_(k-1) > 1 for example, and get k < 4.2,
T_k is bigger than T_(k-1) for up to k = 4
so the biggest term is T_k which is T_4.


EDIT: from her next post I don't think this is what she's looking for

 

[psych_girl]

A.) Give an example.

B.) how can you tutor 2U/3U?

I mean the harder 3U question on Binormal, tutor is ppl who can be confused sometimes, dont think tutor is a superman!

 

[dansk1er]

(a) Putting x = 1 into the expansion of (1 + x)ⁿ gives: c0 + c1 + c2 + ... + cn = 2ⁿ _____ (1)

Differentiating the expansion of (1 + x)ⁿ, and then putting x = 1 gives:
c1 + 2 * c2 + 3 * c3 + ... + n * cn = n * 2^n-1 _____ (2)

(1) + (2) gives c0 + 2 * c1 + 3 * c2 + ... + (n + 1) * cn = (n + 2) * 2^n-1

(b) Integrating the expansion of (1 + x)ⁿ, and evaluating the constant by putting x = 0, gives:

c0 * x + c1 * x² / 2 + c2 * x³ / 3 + ... + cn * xⁿ⁺¹ / (n + 1) = [(1 + x)ⁿ⁺¹ - 1] / (n + 1)

Integrating this a second time, again evaluating the constant by putting x = 0.

The required sum can then be found by putting x = 1.

Reading a solution that was posted before i was even born... hope you're doing well