Avi's challenge 3

Probability Level 5

If the coefficient of $x^n$ in expansion of $\dfrac1{(1-x)(1-2x)(1-3x) }$ Is $[a^{(n+x)} -b^{(n+y)} +c]/k$ , find $a+b+x+y+c+k$ .

2 solutions

Avi Solanki
Mar 7, 2016

By resolving into partial fraction

1/2(1-X) -4/(1-2x) +9/2(1-3x) .

=1/2 (1-x)^-1 -4(1-2x)^-1 + 9/2 (1- 3x)^-1

=1/2(1+x+..........x^n...) -4(1+2x+(2x)^2 +.....) + 9/2(1+3x+......(3x)^n...)

=>coefficient of x^n =1/2.[ 1-8.2^n + 9.3^n]

=1/2[ 1- 2^(n+3) + 3^(n+2) ] .

Hence.
A+b+c+x+y+k =13

it was a nice one go on framing such creative questions.

aryan goyat - 5 years, 3 months ago

Thank u aryan . I would love to know another method to solve this question if u could provide an innovative solution

Also try the question "dude thats a huge sum"

avi solanki - 5 years, 3 months ago

same solution here... for me the most time taking part is to resolve into partial fraction

展豪張 - 5 years, 3 months ago

Try this problem as well "14 balls only"

avi solanki - 5 years, 3 months ago

Aryan Goyat
Mar 7, 2016

the above can also be written as (1+x+x^(2)+x^(3)............)(1+2x+4x^(2)+8x^(3).......)(1+3x+9x^(2).....)

the coefficient of x^(n)

can be seen as

a+b<=n

sum of 3^(a)2^(b)

3^(0){2^(n)+2^(n-1)+--------+1}

3^(1){2^(n-1)+2^(n-2)+--------+1}

..................

3^(n){2^(0)}

simplying using sum of GP and using a^(n)b^(0)+a^(n-1)b^(1)+..........+b^(n)={a^(n+1)-b^(n+1)}/(a-b)