 # Sum  to  n    terms      3.7.11.15+7.11.15.19+................... Dont post the full solution,Maybe Hints(Like the method )Should we use factorial things in here???

11 years ago

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

11 years ago

Do u know the Vn method ??

11 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

11 years ago

hi swapnil

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

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

11 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

11 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) + 3.7.11.15] ------- (Required Answer).

Regards,

11 years ago

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

S = a1a2a3a4a5....ar + ar+1ar+2ar+3ar+4....a2r +    ..............  + an-r+1an-r+2an-r+3......an

Tn = an-r+1an-r+2an-r+3......an

Vn = anan+1an+2......an+r-1an+r

(Take an extra Last Term in Vn)

Vn-1 = an-1anan+1......an+r-2an+r-1

Vn - Vn-1 = anan+1an+2......an+r-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]

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 .

11 years ago

@ Ashwin Sir,

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)

11 years ago

Dude,

Enna idhu Sir lam... :P

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.

Regards,

11 years ago

Pingilikka pillaaapi ,Summa said.

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

11 years ago

Dude,

Then please get comfortable with it.

It is really easy.

Regards,

11 years ago

Periya manishan,Neenga solitengala.......Ill get comfortable  :) And also,I have send u sme mails in gmail......Pls look at it

11 years ago

Dude,

Done!!

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

Wish you all the best.

Regards,

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