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# 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???

9 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

9 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

9 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,

9 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 .

9 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)

9 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,

9 years ago

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

9 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,

9 years ago

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

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