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Important Concepts and Formulas - Sequence and Series

 



Comments(19)


priya 04 Apr 2014 10:22 AM
the sum of first 'n' terms of the series 1/2+3/4+7/8+15/16+......
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Jay 05 Apr 2014 2:53 PM
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 2:43 AM
liked the page
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Wilson Wande 13 Mar 2014 2:42 AM
so educative an mathematically enriching
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Warda Sheikh sheikh 16 Feb 2014 12:55 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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Dev 16 Feb 2014 1:24 PM
Let the sides be (9-d), 9,  (9+d) where (9+d) is the  hypotenuse

(9+d)2 = (9-d)2 + 92
92 + 18d + d2 = 92 - 18d + d2 + 92
36d = 81
d = 81/36 = 9/4 = 2.25

ie, sides are (9-2.25), 9,  (9+2.25) or 6.75,  9,  11.25

(clearly this is an AP with 6.752 + 92 = 11.252)
hypotenuse = 11.25 cm

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M.jeevan nayak 05 Feb 2014 3:31 AM
for this solution you can use p^((n^2)-1/n^2)
       
      where n= how many no. you taken
                 p=3
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sreelesh 29 Jan 2014 4:29 AM
if a, b, c are in G.P
a) a(b2+a2) = c(b2+c2)
b) a(b2+c2) = c(a2+b2)
c) a2(b+c) = c2(a+b)
d) None of these
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Jay 30 Jan 2014 12:34 PM
if a, b, c are in G.P, b2 = ac

a(b2+c2) = a(ac + c2) = a2c + ac2
c(a2+b2) = c(a2 + ac) = a2c +ac2

Hence a(b2+c2) = c(a2+b2)
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sudha 29 Jan 2014 3:31 AM
if an a.p sn=3n2+5n then ak =164 find the value k

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