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Sum  to  n    terms    


Dont post the full solution,Maybe Hints(Like the method )


Should we use factorial things in here???



3 years ago


Answers : (14)


The biggest hint to theis problem tht the question can be done using Vn method. Search it. Try it

3 years ago

Do u know the Vn method ??

3 years ago

At Swapnil Saxena : Thanks for the hint.How do u know these kinda of methods?

V-n method

V-n method is used to solve the series summation problem in the fom of :


1> 1/(a1*a2)  +  1/(a2*a3)  +  1/(a3*a4)  +........ upto n terms


2>(a1*a2)  + (a2*a3)  +  (a3*a4)  +.............upto n terms


where a1,a2,a3 ..... are in AP with certain common difference


3 years ago

hi swapnil

this question cannot be done by VN method!!!!

see and guess why this question is not done by VN method

3 years ago

@ Samarth : No Idea??? Why... Tell me. Is there is something wrong. This is wht u told me.

@ Sathyaram : Good Friendship, attempts, courage, and hard labour always pays...

@ Jit: Sathyaram is right... but I wanna further extend . Not just the series like tht but like

a1a2a3a4a5+ a2a3a4a5a6 +a3a4a5a6a7 .... or 1/a1a2a3a4a5+ 1/a2a3a4a5a6 + 1/a3a4a5a6a7 ....  like difficulties can be solved by this method  

3 years ago

Hi Guys,


This question can definitely be solved using the Vn method.

S = Σ(4r-1)(4r+3)(4r+7)(4r+11), r=1,2,3,....n.

So Tr = (4r-1)(4r+3)(4r+7)(4r+11)

and let Vr = (4r-5)(4r-1)(4r+3)(4r+7)(4r+11)

hence, Vr+1 = (4r-1)(4r+3)(4r+7)(4r+11)(4r+15)


So Vr+1 - Vr = 20xTr.

Hence Σ(Vr+1 - Vr) = 20*S

Hence S = (1/20)*[Vn+1 - V1] = (1/20)*[(4n-1)(4n+3)(4n+7)(4n+11)(4n+15) +] ------- (Required Answer).



Ashwin (IIT MadraS).

3 years ago

Agree with swapnil. Lemme try to generalize it for u.


S = + ar+1ar+2ar+3ar+4....a2r +    ..............  +


Tn =


Vn =

(Take an extra Last Term in Vn)


Vn-1 =


Vn - Vn-1 = [an+r - an-1 ] = Tn[an+r - an-1 ] = Tn [{a1 + (n+r-1)d} - {a1+ (n-2)d}] = Tn [(r+1)d]


or, Tn = 1/[(r+1)d] * [Vn - Vn-1]


Now put n=1,2,3,....,n and add.


T= 1/[(r+1)d] * [V1 - V0]

T= 1/[(r+1)d] * [V2 - V1]

T= 1/[(r+1)d] * [V3 - V2]





T= 1/[(r+1)d] * [Vn - Vn-1]


S = 1/[(r+1)d] * [Vn - V0]


From here you can find the sum easily.


Similarly for the sum of reciprocals, Take one term less for Vn .

3 years ago

@ Ashwin Sir,

Thanks for ur answer and ur answer is correct...


But, How could u assume that "let Vr = (4r-5)(4r-1)(4r+3)(4r+7)(4r+11)"?


And where did (4r-5) go in this equation? :

Vr+1 = (4r-1)(4r+3)(4r+7)(4r+11)(4r+15)

3 years ago



Enna idhu Sir lam... :P

Seri adhe vidu.


Vr is assumed based on the Tr.

In Vr we assume one more term in the start.


If Vr = (4r-5)(4r-1)(4r+3)(4r+7)(4r+11)

then replace r by r+1 in this to get Vr+1.

Hence you get Vr+1.



Ashwin (IIT Madras).

3 years ago

Pingilikka pillaaapi ,Summa said.


Can u do this without Vn method ?I am kinda new to this and not feelin comfortable...

3 years ago



Then please get comfortable with it.

It is really easy.



Ashwin (IIT Madras),

3 years ago

Periya manishan,Neenga solitengala.......Ill get comfortable  :)  Laughing  


And also,I have send u sme mails in gmail......Pls look at it

3 years ago




Hope that you take those mail replies seriously. Now you prepare and do your Boards well.


You can think about your 11th Std after your exams :).

Wish you all the best.



Ashwin (IIT Madras).

3 years ago

Oh k.Thnks.I have sent u  reply....Pls look at it and dont forget to c mokkais ! Tongue out

And in Biology CBSE,

I dont understsnd these mendels cross breeding,dominant,recessice...Could u just help me out... :)

3 years ago

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