IIT JEE , AIEEE Forum - Mathematics , Algebra

divij_wadhawan

1/1.2.3 + 1/3.4.5 + 1/5.6.7 +........... upto n terms

the nth term of the sequence comes out to be-->> 1/(2n-1)(2n)(2n+1)

then to find the sum uptil nth term we

1/(2n-1)(2n)(2n+1) =

1/((8n^3)- 2n)

please tell me how to do it furthur ......

and also tell me generally how to solve sum of sequence when the nth term is in denominator like the above questn

ankur.kkhurana

hey buddy is ans is 1/4 - 1/(2n)(2n+1)

abhishekarora

u can do it only by using V_nmethod.

sivateja_abcdefg

is the ans is ln2-1/2

sivateja_abcdefg

sol:

Tn=1/2(2n-1)-2/2n+1/2(2n+1) [on applying partial gractions]

=1/2[1/(2n-1)-1/2n]-1/2[1/2n-1/(2n+1)]

summating upto infinity the terms in brackets are log expansions

after u will get it as ln2-1/2

hope u got that

nigitha_17

s_n=1/a₁a₂......a_n+1/a₂a₃..........a_n-1+....................

then

s_n=1/(r-1)d[1/a₁a₂......a_r-1-1/a_n+1a_n+2..........a_n+r-1-..................]

d=common diff

r=number of factors in denominator

IF U DONN FOLLOW THE FORMULA JUS INFORM ME I'LL GIVE U AN EXAMPLE

khyatigupta

hey i hv got this ans-

Sn=n(n+3)/3(n+1)(n+2)

i have solved it by Vn method.i will tell u the solution if my ans wld be right.

ankur.kkhurana

hey is my answer right .If so i will post solution

Avinash_Bhat

All of you have tried very well. But I don't think any of these are right. I will explain :

Given series is : 1/1.2.3 + 1/3.4.5 + 1/5.6.7 + ................

So : nth term, Tn = 1 / (2n - 1)(2n)(2n + 1)

This is to be split into 3(partial fractions). The method is as follows :

Let 1 / (2n - 1)(2n)(2n + 1) = A/(2n - 1) + B/(2n) + C/(2n + 1)

ie, 1 = A(2n)(2n + 1) + B(2n - 1)(2n + 1) + C(2n)(2n - 1)

Put n = 0 to get B = -1

Put n = -1/2 to get C = 1/2

Put n = 1/2 to get A = 1/2

SO : Tn = 1/2(2n - 1) - 1/(2n) + 1/2(2n + 1)

Required sum of the series :

Tn = { 1/2(2n - 1) - 1/(2n) + 1/2(2n + 1) }

= { 1/2(2n - 1) - 1/(4n) } - { 1/(4n) - 1/2(2n + 1) }

= 1/2( 1 - 1/2 + 1/2 - 1/4 + ........ ) - 1/2(1/2 - 1/3 + 1/4 - 1/5 + ........ )

= 1/2 (log 2) + 1/2 (log 2 - 1)

= 1/2 (log 2) + 1/2 (log 2) - 1/2

= log 2 - 1/2

So, What do you say ???

Avinash_Bhat

Some extras................

1) a₁ , a₂ , a₃ , a₄ , ............ a_n-1 , a_n , a_n+1 in A.P.

Then : 1/a₁a₂ + 1/a₂a₃ + 1/a₃a₄ + ................ + 1/a_na_n+1 = n / a₁a_n+1

2) a₁ , a₂ , a₃ , a₄ , ............ a_n-1 , a_n , a_n+1 in A.P.

Then : 1/

a₁+

a₂ + 1/

a₂+

a₃ + 1/

a₃+

a₄ + ................ + 1/

a_n+

a_n+1 = n /

a₁+

a_n+1

3) a₁ , a₂ , a₃ , a₄ , ............ a_n-1 , a_n , a_n+1, a_n+2 in A.P.

Sn = 1/a₁a₂a₃ + 1/a₂a₃a₄ + 1/a₃a₄a₅ + ............. nth term, Tn = 1/a_na_n+1a_n+2

Let V_n = 1/a_n+1a_n+2

V_n-1 = 1/a_na_n+1

.

V₁ = 1/a₂a₃

V₀ = 1/a₁a₂

V_n-1 - V_n= kTn ; V_n-2 - V_n-1 = kT_n-1 ................. ; V₀ - V₁= kT₁

So : Sn = (V_o - V_n) / k

Avinash_Bhat

SO ............ HOW ARE THEY ????????

prathima

can u plz xplain how did u get log2 ????????????

ALSO CAN U PLZ XPLAIN THE FORMULAS GIVEN BY U ????

PLZ I WOULD BE VERY THANK FULL TO U IF U TAKE A LITTLE PAIN IN EXPLAINING THIS

PLZ ZZZZZZZZZZZZZZZZZ XPLAIN

abhijeet_0201

its a formula that
log(1+x)=x-x^2/2+x^3/3-x^4/4.............
hope u understood!!!!!!!!!!!!!!!

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