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???

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

Do u know the V_n method ??

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

hi swapnil

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

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

@ 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  

Hi Guys,

This question can definitely be solved using the V_n method.

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

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

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

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

So V_r+1 - V_r = 20xT_r.

Hence Σ(V_r+1 - V_r) = 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,

Ashwin (IIT MadraS).

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

S = a₁a₂a₃a₄a₅....a_r + a_r+1a_r+2a_r+3a_r+4....a_2r +    ..............  + a_n-r+1a_n-r+2a_n-r+3......a_n

T_n = a_n-r+1a_n-r+2a_n-r+3......a_n

V_n = a_na_n+1a_n+2......a_n+r-1a_n+r

(Take an extra Last Term in V_n)

V_n-1 = a_n-1a_na_n+1......a_n+r-2a_n+r-1

V_n - V_n-1 = a_na_n+1a_n+2......a_n+r-1 [a_n+r - a_n-1 ] = T_n[a_n+r - a_n-1 ] = T_n [{a₁ + (n+r-1)d} - {a₁+ (n-2)d}] = T_n [(r+1)d]

or, T_n = 1/[(r+1)d] * [V_n - V_n-1]

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

T₁ = 1/[(r+1)d] * [V₁ - V₀]

T₂ = 1/[(r+1)d] * [V₂ - V₁]

T₃ = 1/[(r+1)d] * [V₃ - V₂]

.

.

.

.

T_n = 1/[(r+1)d] * [V_n - V_n-1]

______________________________

S = 1/[(r+1)d] * [V_n - V₀]

From here you can find the sum easily.

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

@ Ashwin Sir,

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

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

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

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

Dude,

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.

Regards,

Ashwin (IIT Madras).

Pingilikka pillaaapi ,Summa said.

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

Dude,

Then please get comfortable with it.

It is really easy.

Regards,

Ashwin (IIT Madras),

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

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

Dude,

Done!!

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.

Regards,

Ashwin (IIT Madras).

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... :)