Finding and proving a formula for 3+7+11+...+(4n-1) by induction

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Demonoid
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I'm a little stuck here...

I need to write this in the summation notation, and then find and prove a formula in terms of n, using induction :3+7+11+...+(4n-1)

I know that the summation notation is
n
+---
\
/ 4i-1
+---
i=1

but I have no idea how to actually find the formula for the sum and then prove it.

3+7=10
3+7+11=21
3+7+11+15 =36
3+7+11+15+19 =55

I don't see a pattern in the summation, so I can't figure it out.Any help will be greatly appreciated !
 

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This is an arithmetic sequence, you should know a formula. If not it should be easy to split the terms into two sums, each having easy formulas.
 
Jarle said:
This is an arithmetic sequence, you should know a formula. If not it should be easy to split the terms into two sums, each having easy formulas.

a little bit more explanation with sample solution will be helpful. What you said doesn't help a bit. I know that it is an arithmetic formula (4n-1). The sum on the other hand is not.
 
Jarle said:
This is an arithmetic sequence, you should know a formula. If not it should be easy to split the terms into two sums, each having easy formulas.

First look at the sum from 1 to n of 4i.

This is:

4 + 8 + 12 + 16 + 20 + ... + 4n

Now, notice how the sum of the 1st and the n-th terms is 4+4n = 4*(n+1).
Notice how the sum of the 2nd and the (n-1)-th terms is 8+4*(n-1) = 4*(n+1)
Notice how the sum of the 3rd and the (n-2)-th terms is 4*(n+1)
Notice how the sum of the 4th and the (n-3)-th terms is 4*(n+1)

If n is even, then the you can see how the above sum must be (n/2)*4*(n+1).

This equals 2*n*(n+1). We can see, however, that this actually works for all n, even or odd, in the following way:

If n=1, then 2*n*(n+1) = 4, which is correct. Now, let n=k, where k is some integer. Consider what happens when when change from n=k to n=k+1. Then the above expression goes from 2*k*(k+1) to 2*(k+1)*(k+2). But notice, this second expression is exactly 4*(k+1) more.

In other words, [2*(k+1)*(k+2)] - [2*k*(k+1)] = 4*(k+1). But 4*(k+1) is exactly the amount we add when we go from n=k to n=k+1. Thus the expression 2*n*(n+1) must be correct for any n.

This type of sum is called an arithmetic sum. The sum from i=1 to n of i equals n*(n+1)/2. This is something you should remember, or else remember how to derive. Once you know this, it is easy to find an expression for any other arithmetic sum. For example, the sum from 1 to n of a*i - b will simply equal a*(n*(n+1)/2) - n*b. But you should not memorize this. You should memorize the manipulations you are allowed to do with series and sequences.

The sum from 1 to n of a*f(i), where f is a function of i, and 'a' is a constant, will always equal 'a' times the sum from 1 to n of f(i).

The sum from 1 to n of f(i) + g(i), where f and g are a functions of i, will always equal [the sum from 1 to n of g(i)] plus [the sum from 1 to n of f(i)].

Specifically, the sum from 1 to n of f(i) + b, where f is a functions of i, and 'b' is a constant, will always equal n*b plus the sum from 1 to n of f(i).
 
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Demonoid said:
a little bit more explanation with sample solution will be helpful. What you said doesn't help a bit. I know that it is an arithmetic formula (4n-1). The sum on the other hand is not.

What I meant was that you probably have learned a formula for summing an arithmetic sequence. Do you know the formula for the sum [itex]\sum_{k=1}^n k[/itex] ? If so, split the sum as this: [itex]\sum_{k=1}^n 4k-1= 4\sum_{k=1}^n k+-\sum_{k=1}^n 1[/itex].

[itex]\sum_{k=1}^n k = \frac{k(k-1)}{2}[/itex] by the way. You probably should show that by induction if this will be the means of your proof.