RadiationX said:
I have a vauge idea of what these rules mean:
1. \sum^n_{i=1}c=cn
2. \sum^n_{i=1}i=\frac{n(n+1)}_{2}
3. \sum^n_{i=1}i^2=\frac{n(n+1)(2n+1)}_{6}
4. \sum^n_{i=1}i^3=\frac{n^2(n+1)^2}_4
are these rules saying that if i have an expression, (i) raised to a power from 1 to 3 then i use the above rules? How would you solve the following problem?
\sum^n_{i=1}\frac{(i + 1)}_{n^2} for n = 10,100,1000,10000
i know that the answer reduces to \frac{(n + 3)}_{2n}
and you make some substitutions, but how is this done? any help would be appreciated. It's three am and I'm still trying to figure this out!
The summation symbol is
the most powerful mathematical symbol there is, so you need to be certain that you do understand its usage. Its usage must be consistent with the axioms of algebra. With that in mind, consider the first equation above:
\sum^n_{i=1}c=cn
The left hand side of this equation is a "shorthand" for the following longer expression:
C1+C2+C3+C4+C5+... +Cn where C1=C2=C3=...=Cn=C
Now, use the distributive law of algebra to factor out the C to get this:
C(1+1+1+1+...+1)
There are n number ones inside the parenthesis, so that you get this after factoring out the C:
C times n = nc
So that the LHS is equivalent to the RHS, by an application of the distributive law of algebra.
Your number two is far more interesting.
\sum^n_{i=1}i=\frac{n(n+1)}_{2}
The LHS uses the summation symbol as a shorthand for the following longer expression:
1+2+3+4+5+... +n
Which is nothing more than the sum of the first n natural numbers. In order to understand the formula on the RHS, consider this:
1+2+3+4+5
5+4+3+2+1
6+6+6+6+6
= 5*6
So that 1+2+3+4+5 = 5*6/2
So that in general:
1+2+3+...+n = n*(n+1)/2
Which is the RHS by ad hoc reasoning.
Now, without deriving number three above, i challenge you to prove that it is true using mathematical induction. You will find that it is true, indeed the LHS is numerically equivalent to the RHS.
All this having been said, consider evaluating the following expression:
\sum^n_{i=1}(6i^2 -4i + 7)
Using the axioms of algebra, the previous expression is numerically equivalent to the following expression:
\sum^n_{i=1}6i^2 - \sum^n_{i=1}4i + \sum^n_{i=1}7
To find the final expression, you will need to know the following, preferably a priori:
1. \sum^n_{i=1}6i^2 = 6\sum^n_{i=1}i^2= n(n+1)(2n+1)
2. \sum^n_{i=1}4i = 4\sum^n_{i=1}i= 2(n)(n+1)
3. \sum^n_{i=1}7=7n
Your final formula will not have the dummy variable i in it, rather it will be a function of n, where n is an element of the natural numbers.
Then if you want to find out what the sum is for n = 10,100,1000,10000, i suggest you buy a pocket calculator for $20, and do the final multiplication. To check your answer, write a computer program with the summation formula in it, so that the computer actually adds, and then wait to see if the answer after long addition, is equivalent to the answer you got on your calculator.
Just so you are fully aware, the axioms of algebra are being used to write these equations. In other words, consider something like this:
\sum^n_{i=1}6i^2
The previous expression is a shorthand for the following expression:
6(1^2)+6(2^2)+6(3^2)+... + 6(n^2)
It is only after an application of the distributive law of algebra, that we can say the previous expression is numerically equivalent to the following 'different looking' expression:
6(1^2+2^2+3^2+...+n^2)
And then we can use the summation shorthand to justify saying that the previous expression is numerically equivalent to the following expression:
6\sum^n_{i=1}i^2
So now let us consider your question, you want to tackle the following SUM:
\sum^n_{i=1}\frac{(i + 1)}_{n^2}
The first thing to realize, is that as the indicial variable i is varying, the number n is remaining constant. Since it is a constant, you can factor it out of the summation symbol (this is justified by the distributive axiom of algebra), so that you can write the following equation:
\sum^n_{i=1}\frac{(i + 1)}_{n^2}= 1/n^2 \sum^n_{i=1}(i + 1)
And now you need to know this:
\sum^n_{i=1}(i + 1) =\sum^n_{i=1}i + \sum^n_{i=1}1 = n(n+1)/2 + n
Now, get a common denominator of 2.
In other words:
n(n+1)/2 + n = n(n+1)/2+2n/2= [ n(n+1)+ 2n ]/2 = n[(n+1) +2]/2 = n(n+3)/2
Now, you can pull everything together to write this:
\sum^n_{i=1}\frac{(i + 1)}_{n^2} = 1/n^2 * n(n+3)/2 = (n+3)/2n
as desired.
