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Important Concepts and Formulas - Sequence and Series

 



Comments(24)


seenu 06 Sep 2014 7:40 PM
i like maths and i love maths
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Harshit 04 Sep 2014 1:01 PM
How to you find the sum of a series if the difference of terms of the series are in ap.
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somesh chandra 07 Jul 2014 5:28 PM

Q. find the ans srise

0,8,54,192,500,?

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Jay 10 Jul 2014 12:10 AM
The pattern is n4 - n3

it goes as 0, 8, 54, 192, 500, 1080, 2058
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vinita 23 May 2014 3:40 PM
i like it
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priya 04 Apr 2014 7:52 PM
the sum of first 'n' terms of the series 1/2+3/4+7/8+15/16+......
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Jay 06 Apr 2014 12:23 AM
Let Sn = 1 + 3 + 7 + 15 + ...  + Tn

Sn = 1 + 3 + 7 + 15 + ...   + Tn                  -----------(1)
Sn =       1 + 3 +  7  + ...   + Tn-1 + Tn        ----------(2)

(1)-(2)=>    0 = 1 + 2 + 4 + 8 + ... (Tn - Tn-1)  - Tn
=> Tn = 1 + 2 + 4 + 8 + ... (Tn - Tn-1)
=> Tn = 1 + 2 + 22 +  23 + ........... = 1(2n-1)/(2-1)  = (2n-1)     [sum of n terms in GP]


i.e., nth term in the series  1 + 3 + 7 + 15 + ...  can be written as (2n-1)   


Hence,
$MF#%\dfrac{1}{2}+\dfrac{3}{4}+\dfrac{7}{8}+\dfrac{15}{16}+... \text{n terms}\\\\
\dfrac{1}{2}+\dfrac{3}{2^2}+\dfrac{7}{2^3}+\dfrac{15}{2^4}+... \dfrac{2^n-1}{2^n}\\\\
= \sum_{r=1}^n\left(\dfrac{2^r-1}{2^r}\right)\\\\
\sum_{r=1}^n\left(\dfrac{2^r-1}{2^r}\right) = \sum_{r=1}^n\left(1 - \dfrac{1}{2^r}\right) \\\\= \sum_{r=1}^n \left(1\right) - \sum_{r=1}^n\left(\dfrac{1}{2^r}\right)\\\\
= n - \dfrac{\dfrac{1}{2}\left[1 - \left(\dfrac{1}{2}\right)^n\right]}{1 - \dfrac{1}{2}}\\\\
= n -  \dfrac{\dfrac{1}{2}\left[1 - \left(\dfrac{1}{2}\right)^n\right]}{\dfrac{1}{2}}\\\\
= n -   \left[1 - \left(\dfrac{1}{2}\right)^n\right]\\\\
= n -  1 + \left(\dfrac{1}{2}\right)^n$MF#%
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Wilson Wande 13 Mar 2014 12:13 PM
liked the page
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Wilson Wande 13 Mar 2014 12:12 PM
so educative an mathematically enriching
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Warda Sheikh sheikh 16 Feb 2014 10:25 PM
the base of a right angel triangle is 9cm and the sides of the triangle are in A.P.finf hyptenuse.



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