## Important Concepts and Formulas - Sequence and Series

Comments(24)

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.

somesh chandra
07 Jul 2014 5:28 PM

Q. find the ans srise

0,8,54,192,500,?

Jay
10 Jul 2014 12:10 AM

The pattern is n

it goes as 0, 8, 54, 192, 500, 1080, 2058

^{4}- n^{3}it goes as 0, 8, 54, 192, 500, 1080, 2058

priya
04 Apr 2014 7:52 PM

the sum of first 'n' terms of the series 1/2+3/4+7/8+15/16+......

Jay
06 Apr 2014 12:23 AM

Let S

S

S

(1)-(2)=> 0 = 1 + 2 + 4 + 8 + ... (T

=> T

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#%

_{n}= 1 + 3 + 7 + 15 + ... + T_{n}S

_{n}= 1 + 3 + 7 + 15 + ... + T_{n}-----------(1)S

_{n}= 1 + 3 + 7 + ... + T_{n}_{-1}+ T_{n}----------(2)(1)-(2)=> 0 = 1 + 2 + 4 + 8 + ... (T

_{n - }T_{n}_{-1) - }T_{n}=> T_{n}= 1 + 2 + 4 + 8 + ... (T_{n - }T_{n-1})=> T

_{n}= 1 + 2 + 2^{2}+ 2^{3}+ ........... = 1(2^{n}-1)/(2-1) = (2^{n}-1)*[sum of n terms in GP]*

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

^{n}-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#%

_{}
Wilson Wande
13 Mar 2014 12:12 PM

so educative an mathematically enriching

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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