## Yahoo! 3 years ago Find the n th term and sum of n terms of the series 5 + 8 + 17 + 44 ..........

1. dpflan

Ah, this is a nice series. $[5,8,17,44]$

2. experimentX

find the difference .. note the difference in the difference is a multiple of 3

3. Yahoo!

Yes the first order difference is in Gp

4. dpflan

and on...so do as @experimentX suggests, first examine any sort of relationship that could exist, you see that each termer is greater than the next

5. dpflan

I suggest that you show us the values you have for the difference between the numbers in the series

6. dpflan

Once, you have those differences, see if there is a pattern or relationship among them

7. zepp

Looks like 3 to some power :D

8. experimentX

tn+1 = tn + 3^(n-1)

9. dpflan

term1 = 5 term2 = 8 term2 - term1 = 3 so term2 = term1 + 3 term3 = 17 term3 - term2 = 9 so term3 = term2 + 9 term4 = 44 term5 - term4 = 27 so term4 = term3 + 27 term5 = ...

10. experimentX

don't know closed form at the moment!!

11. dpflan

So @Yahoo! How are you doing with this?

12. Yahoo!

Ar^n + b1n^k-1 + b2n^k-2............bk

13. dpflan

Hmm, can you explain the terms please?

14. Yahoo!

it is the general formula lol1

15. sami-21

ok look at the sequence first term is (3^1+7)/2=5 second is (3^2+7)/2=8 third is (3^3+8)/2=17 can you now tell me general formula ?

16. experimentX

i guess we should use summation rule |dw:1343147827410:dw|

17. mahmit2012

|dw:1343147906405:dw|

18. mahmit2012

|dw:1343147929358:dw|

19. experimentX

not sure that works!

20. experimentX

|dw:1343147968035:dw| something like that!! just try to manipulate and fit in it ...

21. mahmit2012

|dw:1343147962155:dw|

22. experimentX

|dw:1343148038487:dw|

23. mahmit2012

|dw:1343148023475:dw|

24. amistre64

i like to get the first term to 1 alot of times 5 + 8 + 17 + 44 -4 -4 -4 -4 ----------------- 1 + 4 + 13 + 40 just helps me with the inbetween addings or if we up it by +1 5 + 8 + 17 + 44 +1 +1 +1 +1 ------------------- 6 + 9 + 18 + 45

25. amistre64

factor out a 3 3( 2 + 3 + 6 + 15 + ...) all ways to se it differently

26. experimentX

|dw:1343148174027:dw|

27. experimentX

i hope it works for other terms too .. let's see for sum

28. Yahoo!

@experimentX i didnt understand!!

29. experimentX

|dw:1343148311125:dw|

30. experimentX

that's just some manipulation .. typing in latex really takes lot of time

31. experimentX

$t_n = t_{n-1} + 3^{n-1}$ this is the general rule $t_{n} = t_{n-2} + 3^{n -2} + 3^{n-1}$ continue this way and you will get $t_{n} = t_1 + 3 + 3^3 + 3^4+ .. + 3^{n -2} + 3^{n-1} = 5 + 3{ 3^{n-1} - 1 \over 3-1}$ which is requited formula

32. experimentX

for summation $\sum_{n=1}^k {5 + {3 \over 2} (3^{n - 1 } - 1)} =\sum_{n=1}^k{7/2} + 3^n = k {7 \over 2}+ 3{3^k -1 \over 3-1} = k {7 \over 2}+ {3 \over 2}(3^k -1)$ @Yahoo! do you understand??

33. mahmit2012

|dw:1343149457407:dw|

34. mahmit2012

@experimentX I guess you missed 1/2 .

35. experimentX

thanks @mahmit2012 for correction ... you are sharp!!ii

36. Yahoo!

@mahmit2012 @experimentX i learnt this by using this Ar^n + b1n^k-1 + b2n^k-2............bk

37. Yahoo!

formula!1

38. mahmit2012

You're welcome , and good tutoring you have had. @experimentX

39. Yahoo!

Ar^n + b1n^k-1 + b2n^k-2............bk

40. Yahoo!

can u ssolve this by using Ar^n + b1n^k-1 + b2n^k-2............bk

41. experimentX

you are a good tutor!! most of me is autodidact!!