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its been all morning since ive been trying to understand " arithmetic geometric progression" ...i have read the same theory and the same formulas for finding sum of "n " terms and have tried to understand it and solve questions but i cant, i just dont get it , pls help me,

 

here is a referce site, the theory of this site is exactly the same as my book so im putting it up, pls look it up and help me understand , i dont get the formula and its derivation and even if i try to mug it up , i cant solve the questions with it, one or 2 egs of the questions are :

 

find the sum of n terms of the series:

 

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

 

B ) 1.2^2 + 2.3^2 + 3.4^2 + ......

 

pls help someone, the site is http://www.askiitians.com/iit-jee-progressions-and-series/arithmetic-geometric-progression/

rosala  Nov 11, 2017
 #2
avatar+93876 
+1

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.   surprise

 

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

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

 

 

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

 

------------------------------------------

 

Oh dear, some of my latex has been trucated. surprise sad

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

Melody  Nov 12, 2017
edited by Melody  Nov 12, 2017
 #3
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0

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.

Guest Nov 12, 2017
 #4
avatar+93876 
+1

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

 

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......

Melody  Nov 12, 2017
 #5
avatar+93876 
0

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

Melody  Nov 12, 2017
 #6
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0

This formula given by W/A in the above link of Melody will sum up any number of terms of B above:

So, the sum of the first 100 terms of B above will be:

1/12 n (n + 1) (n + 2) (3n + 5)=1/12[100 x 101 x 102 x 305] =26,184,250.

Guest Nov 12, 2017
 #7
avatar+20153 
+3

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 + ......

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

 

laugh

heureka  Nov 13, 2017
edited by heureka  Nov 13, 2017
edited by heureka  Nov 13, 2017
 #8
avatar+93876 
+2

Thanks Heureka, a great one for me to study :)

Much appreciated :)

Melody  Nov 13, 2017

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