# Find the sum of the series 2 + 3x2 + 4x2^2 + 5x2^3.....[EDITED]

#### Colin67

##### New member

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#### pka

##### Elite Member
The first thing you need is the correct representation of the sum: $$\displaystyle\sum\limits_{k = 2}^n {k\cdot {2^{k - 1}}}$$
If $$S=2+3\cdot 2+4\cdot 2^2+\cdots+n\cdot 2^{n-1}$$ SEE HERE

#### Colin67

##### New member
So the AP is still 2,3,4...

The GP last term is (n-1), so sum is 2(2^(n-1) -1)

Is that correct?

#### Jomo

##### Elite Member
Why would the product be the same as the original? Did you even look at the product?

Is (2+3+4+....)(2+2^2+2^3+...) = 2+ 3*2 + 4*2^2 + 5*2^3 +... ?

Why are you having trouble finding the product? (n/2)(3+n)*2(2^n-1) = n(3+n)*(2^n-1).

Have you consider proving this by induction?

#### Jomo

##### Elite Member
2*2^0+ 3*2^1 + 4*2^2 + 5*2^3 +... + (n+1)2^n

= 2(2^0 + 2^1 + .... + 2^(n-1)) + (1*2^1+ 2*2^2 + 3*2^3 + ... n*2^n)
= 2(2^0 + 2^1 + .... + 2^(n-1)) + 2( 2^0 + 2^1 + ... + 2^(n-1) + (1*2^2 + 2*2^2 +...(n-1)^(n-1))
=....

#### pka

##### Elite Member
So the AP is still 2,3,4...

The GP last term is (n-1), so sum is 2(2^(n-1) -1)

Is that correct?
Did you use the link? $$\sum\limits_{k = 2}^n {k\cdot{2^{k - 1}}} = {2^n}(n - 1)$$
If we could use calculus the proof is easy.

#### Colin67

##### New member
Hi

Yes I looked at Wolfram link. As is clearly apparent I am still not sure how to tackle this problem.

Jomo I did try and expand brackets and got n2^n-n-3+3(2)^n, and no further.

Hence my request for help, especially because i think the basic way I am approaching this is wrong

#### Subhotosh Khan

##### Super Moderator
Staff member
$$\displaystyle S \ = \ \sum_{i = 1}^n \left[(i+1)*2^{i-1}\right]$$

$$\displaystyle \ S \ = 2 + 3*2 + 4*2^2 + ...................... n*2^{n-2} + (n+1)*2^{n-1}$$ ......................................................(1)

$$\displaystyle 2S \ = \ \ \ \ \ 2*2 + 3*2^2 + 4*2^3 + ............................ n*2^{n-1} + (n+1)*2^{n}$$ ....................................(2)

Subtracting (2) from (1)

$$\displaystyle -S \ = \ \ \ \ \ 2 + \left [2 + 2^2 + 2^3 + ............................ 2^{n-1}\right ] - (n+1)*2^{n}$$ ....................................(3)

continue......

#### Colin67

##### New member
Thank you so much for that, after finding the sum of 2+2^2+2^3...,which is 2^n -2, expanding the brackets and cancelling it leads to n2^n.

You multiplied equation 1 by 2 because it was the common ratio.

#### Jomo

##### Elite Member
Jomo I did try and expand brackets and got n2^n-n-3+3(2)^n, and no further.
You missed my point completely. When you multiply (2+3+4+....) and (2+2^2+2^3+...) do you get 2+ 3*2 + 4*2^2 + 5*2^3 +.... If not then multiplying (2+3+4+....) and (2+2^2+2^3+...) would not be helpful as it does not equal the original equation, 2+ 3*2 + 4*2^2 + 5*2^3 +...

Is that clear? Basically you are saying this: I do not know how to add 1+ pi + 12.34 + sqrt (17) so i will multiply 6 and 14, get 84 and be done. The thing is that 6*14 is NOT equal to the sum.