## 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+......

Jay
05 Apr 2014 2:53 PM

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 2:42 AM

so educative an mathematically enriching

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.

Dev
16 Feb 2014 1:24 PM

Let the sides be (9-d), 9, (9+d) where (9+d) is the hypotenuse

(9+d)

9

d = 81/36 = 9/4 = 2.25

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

(clearly this is an AP with 6.75

hypotenuse = 11.25 cm

(9+d)

^{2}= (9-d)^{2}+ 9^{2}9

^{2}+ 18d + d^{2 }= 9^{2}- 18d + d^{2}+ 9^{2}36d = 81d = 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 6.75

^{2}+ 9^{2}= 11.25^{2})hypotenuse = 11.25 cm

^{}
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

where n= how many no. you taken

p=3

sreelesh
29 Jan 2014 4:29 AM

if a, b, c are in G.P

a) a(b

^{2}+a^{2}) = c(b^{2}+c^{2})b) a(b

^{2}+c^{2}) = c(a^{2}+b^{2})c) a

^{2}(b+c) = c^{2}(a+b)d) None of these

Jay
30 Jan 2014 12:34 PM

**if a, b, c are in G.P, b**

^{2}= aca(b

^{2}+c

^{2}) = a(ac + c

^{2}) = a

^{2}c + ac

^{2}

c(a

^{2}+b

^{2}) = c(a

^{2}+ ac

^{}) = a

^{2}c +ac

^{2}Hence a(b

^{2}+c

^{2}) = c(a

^{2}+b

^{2})

sudha
29 Jan 2014 3:31 AM

if an a.p s

_{n}=3n^{2}+5n then a_{k}=164 find the value k
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