Help with Past Paper Qu. (1 Viewer)

[currysauce]

The 2002 Past Paper, qu. 7 b i) and ii)


per haps i should try my best to type it


nCk = ck

(i) Show that

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

(ii) Find the sum

c0/1.2 - c1/2.3 + c2/3.4 - ...+(-1)^n (cn/(n+1)(n+2) Writing the answer as a simple expression in terms of n

 

[QuaCk]

i can't really see the question.
but with the first part it usually means finding the coefficient of a certain term and equating them on both sides.

 

[acmilan]

Perhaps the question looks better like this

ⁿC_k = c_k

Show:

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

(ii) Find the sum

c₀/1.2 - c₁/2.3 + c₂/3.4 - ...+(-1)ⁿ(c_n/(n+1)(n+2))
Writing the answer as a simple expression in terms of n

 

[currysauce]

yeah.... lol


any ideas?

all i can think of is diferentiating the first one and subbing x=1 but then i get stuck

 

[QuaCk]

how do you write in subscript? :S

anyways ....

Show:

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

using (1+x)^n

(1+x)^n = c0 + c1 x + c2 x^2 + ... + cn x^n
multiply all by x
.: x(1+x)^n = xc0 + xc1 x + xc2 x^2 + ... + x cn x^n
= xc0 + c1 x^2 + c2 x^3 + ... + cn x^(n+1)
differentiate with respect to x:
(1+x)^n + nx(1+x)^(n-1) = c0 + 2c1 + ... + (n+1) cn x^n
(1+x)^(n-1) . [(1+x) + nx] = c0 + 2c1 + ... + (n+1) cn x^n
sub in x=1
.: c0 + 2c1 + 3c2 + .. + (n + 1)cn = (n+2)2^(2n-1) as required.



(ii) Find the sum

c0/1.2 - c1/2.3 + c2/3.4 - ...+(-1)n(cn/(n+1)(n+2))
Writing the answer as a simple expression in terms of n

using (1+x)^n

(1+x)^n = c0 + c1 x + c2 x^2 + ... + cn x^n
integrate with respect to x

(1+x)^(n+1) / (n+1) = c0 x + c1 x^2 + ... + cn x^(n+1) / (n+1) + k
find k by subbing x = 0
.: k = 1/(n+1)

.: (1+x)^(n+1) / (n+1) = c0 x + c1 x^2 + ... + cn x^(n+1) / (n+1) + 1/(n+1)
integrate with respect to x again

(1+x)^(n+1)/(n+1)(n+2) = c0 x^2/1.2 + c1 x^3/2.3 + ... + cn x^(n+2)/(n+1)(n+2) + x/(n+1) + t
find t by subbing x = 0
.: t = 1/(n+1)(n+2)

.: (1+x)^(n+1)/(n+1)(n+2) = c0 x^2/1.2 + c1 x^3/2.3 + ... + cn x^(n+2)/(n+1)(n+2) + x/(n+1) + 1/(n+1)(n+2)

and sub in x = -1 ...
you can do the rest :S
gawd that took soo long to write
hope it's legible

 

[currysauce]

thankyou for your time