IIT JEE , AIEEE Forum - Mathematics , Algebra

nano0101

whats the logic to solve these kind of problems like the one given below?

Find sum till n terms:

3.7.11.15+7.11.15.19+........................

shobna

first find d nth term by subtracting d same sum of d series .
ie, 3.7.11.15+7.11.15.19+......
- 3.7.11.15+7.11.15.19+.....
nd then use d summation method. hope its clear.

shobna

its typing mistake. 3 shud be below 7.

catch_arnnie

3.7.11.15+7.11.15.19+........................

just by carefully observing the series we get the general term as follows

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

this question is based on V_n method

sum of series(s) = _{[ r =1 ]}

^[n](4r-1)(4r+3)(4r+7)(4r+11)

To solve this we, multiply the terms in summation with the difference of the next term to the last term & the previous terms to the first term

& divide the RHS with the required constant

sum of series(s) = (1/20) _{[ r =1 ]}

^[n](4r-1)(4r+3)(4r+7)(4r+11) [(4r+15)-(4r-5)]

=> s= (1/20) _{[ r =1 ]}^[n] [(4r-1)(4r+3)(4r+7)(4r+11)(4r+15)-(4r-1)(4r+3)(4r+7)(4r+11)(4r-5)]

putting the values of r & observing the pattern you'll get the ans.

irlmaks

I think i have a better solution

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

=16(4r+3)(4r+7)(4r+11)

=16t_r /4r-1

implies 19t_r+t_r+1=4r.t_r+1- 4(r-1)t_r

telescoping we get 4n.t_n+1-4.0(t₁)=20S_n+t_n+1-t₁

hence S_n=[(4n-1)(4n+3)(4n+7)(4n+11)(4n+15)+3.7.11.15]/20

DONT HESITATE TO ASK FOR ELOBERATION OF MY SOLUTION IF U THINK MY SOLUTION IS OVER ELAGENT

catch_arnnie

my solution will give the same ans. but its quite long & complicated....

irlmaks

each step i used will be highly useful for solving problems

and mainly that tr-t(r-1)

and telescoping as well known is an excellent tool

magiclko

catch_arnie is correct, such problems are done thru Vn method (as i normally call it)

basically the thing to be done here is, write the rth term of the given series...

then if the series is in the product form , write a term Vr, containing one term more than the given ones, like if the Tr contains 4 terms, then Vr will contain 5 terms....

and if its of quotient form [like 1/ (1.2.3) + 1/(2.3.4)+....], then Vr will contain a term less than Tr, [if Tr =1/ (n+r) (n+r+1)(n+r+2), then Vr =1/ (n+r+1)(n+r+2) ]

in both the cases, on taking difference of Vr and V(r-1), u'll get Tr

and thus on summation it will be left with only two terms.... which can easily be calculated.....

any confusion, do ask again

nano0101

Thank you all !!!!

Ask & Discuss Questions with Community & Experts