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Further Maths DRV question

This is the question

Given P(X=x) = kx for values of x: 1,2,3...n

i) Find k in terms of n
ii) Find E(X)
iii) Find Var(X)

Cheers
I know that 1k+2k+3k...+nk add to 1. But thats as far as I got.
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Original post by BabarTheBast*rd
I know that 1k+2k+3k...+nk add to 1. But thats as far as I got.


Do you know how to sum a series?
1k+2k+3k...+nk = 1 The left hand side can be written as ....
Original post by Muttley79
Do you know how to sum a series?
1k+2k+3k...+nk = 1 The left hand side can be written as ....

Idk Ive never done something like that before
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Use the fact that 1+2+3+4+....+n = n(n+1)/2
Original post by 3pointonefour
Use the fact that 1+2+3+4+....+n = n(n+1)/2

Thats exactly the part I dont understand. I can do everything else and I know it does equal n(n+1)/2 but I can get my head around why
Original post by BabarTheBast*rd
Thats exactly the part I dont understand. I can do everything else and I know it does equal n(n+1)/2 but I can get my head around why


Are you asking for a proof of why the sum of n natural numbers = n(n+1)/2?

The way I tend to do it is consider the "cumulative sum"

So a sequence that goes 1, 1+2, 1+2+3, 1+2+3+4.....

This sequence can be written as 1,3,6,10....

The sum of natural numbers would basically be the nth term of this sequence.

Given that the difference of differences is constant, we can deduce it's a quadratic sequence.

Doing some maths to find out the exact sequence leads to n(n+1)/2
Original post by BabarTheBast*rd
Thats exactly the part I dont understand. I can do everything else and I know it does equal n(n+1)/2 but I can get my head around why


It's an arithmetic series - you should have covered them in Maths A level ...
Original post by 3pointonefour
Are you asking for a proof of why the sum of n natural numbers = n(n+1)/2?

The way I tend to do it is consider the "cumulative sum"

So a sequence that goes 1, 1+2, 1+2+3, 1+2+3+4.....

This sequence can be written as 1,3,6,10....

The sum of natural numbers would basically be the nth term of this sequence.

Given that the difference of differences is constant, we can deduce it's a quadratic sequence.

Doing some maths to find out the exact sequence leads to n(n+1)/2

Oh I see that makes sense now. It is basically the nth term of the triangle numbers. Thanks I understand now.
Original post by Muttley79
It's an arithmetic series - you should have covered them in Maths A level ...

How would you then calculate E(X)
k+4k+9k+16k...+n^2k
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Original post by BabarTheBast*rd
How would you then calculate E(X)


By definition, E[X]=x=1nxP(X=x)=x=1nkx2\displaystyle \mathbb{E}[X] = \sum_{x=1}^n x P(X=x) = \sum_{x=1}^n kx^2

Here you need to know your formula:

12+22+32++n2=16n(n+1)(2n+1)1^2+2^2+3^2 + \ldots + n^2 = \dfrac{1}{6}n(n+1)(2n+1)

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