this is so annoying pls help me

I have never done anything like this before so it was a little difficult for me too.

There are some typo style errors in the online derivation which certainly would have helped confuse you.   

Each term if the product of an AP term and a GP term

The AP is             a,       a+d,             a+2d,     ….       a+(n-1)d     ….

The GP is             b,         br,             br^2,     ……      br^(n-1), …..

The AP-GP is     ab,   (a+d)br,    (a+2d)br^2,   ……  [a+(n-1)d]br^n-1 ….

Now you want the sum to n terms.

\(S_n=ab+(a+d)br \;+(a+2d)br^2 +(a+3d)br^3\;+\; \dots\;+[a+(n-1)d]br^{n-1}\\ \text{multiply both sides by r}\\ rS_n=\qquad \qquad abr \;+\; (a+d)br^2 \;+\; (a+2d)br^3 \;+\; \dots\;\;+[a+(n-2)d]br^{n-1}+\;\; [a+(n-1)d]br^{n}\\ subtract\\ (1-r)S_n=ab+br(a+d-a)+br^2(a+2d-a-d)+...+br^{n-1}[a+(n-1)d-\{a+(n-2)d\}]-[a+(n-1)d]br^{n}\\ (1-r)S_n=ab+br(d)+br^2(2d-d)+...+br^{n-1}[(n-1)d-\{(n-2)d\}]-[a+(n-1)d]br^{n}\\ (1-r)S_n=ab+br(d)+br^2(2d-d)+...+br^{n-1}[(nd-d-nd+2d]-[a+(n-1)d]br^{n}\\ (1-r)S_n=ab+brd+br^2d+...+br^{n-1}d-[a+(n-1)d]br^{n}\\ (1-r)S_n=ab+bdr+bdr^2+...+bdr^{n-1}-[a+(n-1)d]br^{n}\\~\\ (1-r)S_n=ab+[bdr+bdr^2+...+bdr^{n-1}]-[a+(n-1)d]br^{n}\\ \\~\\(1-r)S_n=ab+[sum\;of\;GP]\qquad \qquad -[a+(n-1)d]br^{n}\\ \)

Let's look at the sum of a GP part

\(bdr+bdr^2+...+bdr^{n-1}\\ \text{The first term is bdr and the common ratio is r so}\\ =(bdr)+(bdr)r+...+(bdr)r^{n-2}\\ \text{So the last term is the n-1 term}\\ \text{so in the formula for the sum of a GP r is r, a is replaced with bdr, and n is replaced with n-1}\\ =\frac{bdr(1-r^{n-1})}{1-r}\\ =\frac{bdr(\frac{r-r^n}{r})}{1-r}\\ =\frac{bd(r-r^n)}{1-r}\\ =\frac{bdr(1-r^{n-1})}{1-r}\\ so\\ (1-r)S_n=ab+[sum\;of \;a\;GP]-[a+(n-1)d]br^{n}\\ (1-r)S_n=ab+[\frac{bdr(1-r^{n-1})}{1-r}]-[a+(n-1)d]br^{n}\\ \)

\((1-r)S_n=ab+[\frac{bdr(1-r^{n-1})}{1-r}]-[a+(n-1)d]br^{n}\\~\\ S_n=\dfrac{ab}{1-r}+\dfrac{bdr(1-r^{n-1})}{(1-r)^2}-\dfrac{[a+(n-1)d]br^{n}}{1-r}\\\)

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Oh dear, some of my latex has been trucated.  

I think you should still be able to work it out but let me know if you cannot follow it :/    

B) I do not know the formula for this, but summing up the first 10 terms gives:

∑[ n^2*(n-1), n=2 to 11] =3,850.

\( S_n=\frac{ab}{1-r}+\frac{bdr(1-r^{n-1})}{(1-r)^2}-\frac{[a+(n-1)d]br^{n}}{1-r}\\ \)

find the sum of n terms of the series:

A) 1+ 2x + 3x^2 + 4x^3 +.....

AP=1,  2,   3, ................(n-1) .......     a=1, d=1

GP=1, x, x^2,   ..............x^(n-1)         b=1, r=x

so

\( S_n=\frac{ab}{1-r}+\frac{bdr(1-r^{n-1})}{(1-r)^2}-\frac{[a+(n-1)d]br^{n}}{1-r}\\ S_n=\frac{1*1}{1-x}+\frac{1*1*x(1-x^{n-1})}{(1-x)^2}-\frac{[1+(n-1)*1]1*x^{n}}{1-x}\\ S_n=\frac{1}{1-x}+\frac{x(1-x^{n-1})}{(1-x)^2}-\frac{[1+(n-1)]x^{n}}{1-x}\\ S_n=\frac{1}{1-x}+\frac{x(1-x^{n-1})}{(1-x)^2}-\frac{nx^{n}}{1-x}\\ S_n=\frac{1-nx}{1-x}+\frac{x(1-x^{n-1})}{(1-x)^2}\\ \)

B ) 1.2^2 + 2.3^2 + 3.4^2 + ...   

oh dear, this one is harder......

B ) 1.2^2 + 2.3^2 + 3.4^2 + ......

This is what Wolfram alpha has to say about B

(I'm still thinking about it)

http://www.wolframalpha.com/input/?i=1*2%5E2+%2B+2*3%5E2+%2B+3*4%5E2+%2B+...+%2Bn*(n%2B1)%5E2

find the sum of n terms of the series:

A) 1+ 2x + 3x^2 + 4x^3 +.....

like Euler:

\(\begin{array}{|rcll|} \hline \mathbf{s_n} &\mathbf{=}& \mathbf{1+ 2x + 3x^2 + 4x^3 +\ldots +n x^{n-1}} \quad & | \quad \cdot dx \\ s_n dx &=& dx+ 2xdx + 3x^2dx + 4x^3dx +\ldots +nx^{n-1}dx \quad & | \quad \int{} \\ \int{s_n dx} &=& x+ 2\frac{x^2}{2} + 3\frac{x^3}{3} + 4\frac{x^4}{4} +\ldots +n\frac{x^n}{n} \\ \int{s_n dx} &=& x+ x^2 + x^3 + x^4 +\ldots + x^n = \frac{x-x^{n+1}} {1-x} \\ \int{s_n dx} &=& \frac{x-x^{n+1}} {1-x} \\ \int{s_n dx}&=& \frac{x-x^{n+1}} {1-x} = [x-x^{n+1}][(1-x)^{-1}] \quad & | \quad \text{derivate} \\ s_n &=& [1-(n+1)x^n](1-x)^{-1}+ (x-x^{n+1})(-1)(1-x)^{-2}(-1) \\ s_n &=& \frac{1-(n+1)x^n} {1-x} + \frac{x-x^{n+1}} {(1-x)^2} \\ s_n &=& \frac{1-(n+1)x^n} {1-x} \left(\frac{1-x}{1-x}\right) + \frac{x-x^{n+1}} {(1-x)^2} \\ s_n &=& \frac{[1-(n+1)x^n](1-x)+x-x^{n+1}} {(1-x)^2} \\ s_n &=& \frac{1-x-(n+1)x^n+(n+1)x^{n+1}+x-x^{n+1}} {(1-x)^2} \\ s_n &=& \frac{1 -(n+1)x^n+(n+1)x^{n+1} -x^{n+1}} {(1-x)^2} \\ s_n &=& \frac{1 -(n+1)x^n+[-1+(n+1)]x^{n+1} } {(1-x)^2} \\ \mathbf{s_n} & \mathbf{=}& \mathbf{\dfrac{1 -(n+1)x^n+n x^{n+1} } {(1-x)^2} } \\ \hline \end{array}\)

B ) 1.2^2 + 2.3^2 + 3.4^2 + ......

\(\begin{array}{|rcll|} \hline \mathbf{s_n} &\mathbf{=}& \mathbf{1\cdot 2^2 + 2\cdot3^2 + 3\cdot4^2 +\ldots +n \cdot (n+1)^2 } \\ s_n &=& \sum \limits_{k=1}^{n} k \cdot (k+1)^2 \\ &=& \sum \limits_{k=1}^{n} k \cdot (k^2+2k+1) \\ &=& \sum \limits_{k=1}^{n} (k^3+2k^2+k) \\ &=& \sum \limits_{k=1}^{n} (k^3) + \sum \limits_{k=1}^{n} (2k^2) + \sum \limits_{k=1}^{n} (k) \\ &=& \sum \limits_{k=1}^{n} (k^3) + 2\sum \limits_{k=1}^{n} (k^2) + \sum \limits_{k=1}^{n} (k) \\ && \begin{array}{|rcll|} \hline \sum \limits_{k=1}^{n} (k^3) &=& 1^3+2^3+3^3+ \ldots + n^3 = \left(\frac{n(n+1)}{2}\right)^2 \\ \sum \limits_{k=1}^{n} (k^2) &=& 1^2+2^2+3^2+ \ldots + n^2 = \frac{n(n+1)(2n+1)}{6} \\ \sum \limits_{k=1}^{n} (k) &=& 1+2+3+ \ldots + n = \frac{n(n+1)}{2} \\ \hline \end{array} \\ &=& \left(\frac{n(n+1)}{2}\right)^2 + 2\cdot \frac{n(n+1)(2n+1)}{6} + \frac{n(n+1)}{2} \\ &=& \left(\frac{n(n+1)}{2}\right)\left( \frac{n(n+1)}{2} + \frac23\cdot(2n+1) +1\right) \\ &=& \left(\frac{n(n+1)}{2}\right)\left(\frac{3n(n+1)+4(2n+1)+6}{6}\right) \\ &=& \left(\frac{n(n+1)}{12}\right) \Big( 3n(n+1)+4(2n+1)+6 \Big) \\ &=& \frac{n(n+1)[3n(n+1)+4(2n+1)+6]}{12} \\ &=& \frac{n(n+1)(3n^2+3n+8n+4+6)}{12} \\ &=& \frac{n(n+1)(3n^2+11n+10)}{12} \\ \mathbf{s_n} &\mathbf{=}& \mathbf{\dfrac{n(n+1)(n+2)(3n+5)}{12}} \\ \hline \end{array}\)