Hi guys, can help me solve this series?
S=0,2,6,14,30.... given n=1,2,3,4,5... and d=2
S=0,4,12,28,60.... given n=1,2,3,4,5... and d=3
.
.
.
.
.
d is from 2 onwards
S=?n?d
Printable View
Hi guys, can help me solve this series?
S=0,2,6,14,30.... given n=1,2,3,4,5... and d=2
S=0,4,12,28,60.... given n=1,2,3,4,5... and d=3
.
.
.
.
.
d is from 2 onwards
S=?n?d
Look at the differences between successive numbers in both of your series.
the difference is p*2+2 where p is the previous number for d=2 and p*2+4 for d=3, but how to build a relation for this?
Perhaps by using a "Sum from 1 to i" of some function? Also, think about powers of 2...
Its been a long time since I have done this but I seem to remember that you start of by finding how many 'levels' down the difference between two numbers is 1. E.g.
Then this number is the power of n that you use. Cant remember the rest though. May be way off with this too....Code:0 2 6 14 30
2 4 8 16
2 4 8...
A series is just a way of generating a sequence of numbers. Often, people are concerned with arithmetic or geometric series, where the terms are additive or multiplicative respectively.
Take the first example.
n=1 -> a1 = 0
n=2 -> a2 = 2 = a1 + 2^1
n=3 -> a3 = 6 = a2 + 2^2
n=4 -> a4 = 14 = a3 + 2^3
....
When you have an arithmetic series, you are looking at the difference between each term and the next. In a geometric series you look at the ratio of one term to the next.
Having practically handed you the first one on a plate, you can now write down the general formula (hint: it uses powers of 2) and then have a go at #2, which is pretty much identical. You'll be able to see that for nether of these do you actually need to refer to the previous term.
zaza
hmm, i've never thought of using power 2. I'll try work it out. Thx guys.