Series and Induction help needed alot

rajshah428 said:
okay so i got A = 1/3 B = 1 C = 2/3
so the general expression is 1/3x^3 + x^2 + 2/3x
so it can be written as x^3 + 3x^2 + 2x...........................No
so that is the conjecture.

now for 2c. how do you do it?
Click to expand...

It should be:

\(\displaystyle S_n = \frac{n^3 + 3n^2 + 2n}{3} = \frac{n(n+1)(n+2)}{3}\)

now for 2c. how do you do it?
Click to expand...

This is your problem - tell how you would approach it.

You had defined the sum to be:

\(\displaystyle S_n = \sum_{i=1}^n{i(i+1)} = \sum_{i=1}^n{\left (i^2+i\right )}\)

\(\displaystyle \sum_{i=1}^n i = ??\)

Do not wait for spoon-feeding.
 
Wait so am i supposed to use the method of 'subhotosh' or 'galactus' << induction method for 2c.
And someone on this thread said that 2c is simply adding up the squares and so it should be n(n+1)(2n+1) / 6?
......
and is my working for question 3a corrrect
T_1 = 6
T_2 = 6 + 24 = 30
T_3 = 30 + 60 = 90
T_4 = 90 + 120 = 210
T_5 = 210 + 210 = 420
T_6 = 420 + 336 = 756
^^^^^^^^^^^^^^^^^^^ are all of them correct, someone said they are wrong, btw i havent taken n(n+1)(n+2)
Click to expand...

@ subhotosh isnt S_n = stigma n(n+1)/2 instead of stigma n(n+1)
and i dont know wht n is in the stigma form, so how do i continue
 
rajshah428 said:
Wait so am i supposed to use the method of 'subhotosh' or 'galactus' << induction method for 2c.
And someone on this thread said that 2c is simply adding up the squares and so it should be n(n+1)(2n+1) / 6?
......
and is my working for question 3a corrrect
T_1 = 6
T_2 = 6 + 24 = 30
T_3 = 30 + 60 = 90
T_4 = 90 + 120 = 210
T_5 = 210 + 210 = 420
T_6 = 420 + 336 = 756
^^^^^^^^^^^^^^^^^^^ are all of them correct, someone said they are wrong, btw i havent taken n(n+1)(n+2)
Click to expand...

@ subhotosh isnt S_n = stigma n(n+1)/2 instead of stigma n(n+1)
and i dont know wht n is in the stigma form, so how do i continue
Click to expand...

According to your original post:
2. Consider the series S*n = a*1 + a*2 + a*3 + ... + a*n where a*k is defined as above
Click to expand...

so

\(\displaystyle S_n = \sum_{i=1}^{n}{i(i+1)\)
 
Sometime has passed - I don't know whether the student resolved the issue or not. Just to be complete:

\(\displaystyle \frac{n(n+1)(n+2)}{3} = S_n = \sum_{i=1}^{n}{i(i+1)\)

\(\displaystyle \sum_{i=1}^{n}{i^2} + \sum_{i=1}^{n}{i} = \frac{n(n+1)(n+2)}{3}\)

\(\displaystyle \sum_{i=1}^{n}{i^2} + \frac{n(n+1)}{2} = \frac{n(n+1)(n+2)}{3}\)

\(\displaystyle \sum_{i=1}^{n}{i^2} = \frac{n(n+1)(n+2)}{3} - \frac{n(n+1)}{2} = \frac{n(n+1)}{6} \cdot \left [2(n+2) - 3\right ] = \frac{n(n+1)(2n+1)}{6}\)
 
Hey i have the same problem
for number 5 how do you formulate a conjecture for that series?
 
\(\displaystyle 1^2+2^2+3^2+4^2+5^2+.....\)

\(\displaystyle =1(1)+2(2)+3(3)+4(4)+5(5)+.......\)

\(\displaystyle =[2(1)-1]+[3(2)-2]+[4(3)-3]+[5(4)-4]+........\)

\(\displaystyle 2(1)+3(2)+4(3)+5(4)+...... -(1+2+3+4+....)\)

= original sum - \(\displaystyle \sum_{k=1}^n k\)

This is a solution to 2(c)
 
For Q5,

it can be developed from \(\displaystyle 1(2)(3)4+2(3)4(5)+3(4)5(6)+......\)

\(\displaystyle =1(1+1)(1+2)(1+3)+2(2+1)(2+2)(2+3)+3(3+1)(3+2)(3+3)+...\)

The first terms of each expansion are....

\(\displaystyle 1^4,\ 2^4,\ 3^4,\ 4^4....\)
 
\(\displaystyle S_n= \displaystyle\sum_{i=1}^{n}i^k =\displaystyle\frac{n^k^+^1}{k+1} + \displaystyle\frac{n^k}{2}+\displaystyle\frac{kn^k^-^1}{12}\)
That's what I got for my conjecture for #5.... Not really sure about it though... at all. I said to form a conjecture from the patterns above (the second, third and fourth power series: \(\displaystyle i^2, i^3, i^4\)).
I looked at: http://math2.org/math/expansion/power.htm and found some patterns, while only looking at the second, third and fourth power series.
I also found out that n can't have a power that is negative or a power of zero such as for the power series of 1... Any help... at all? :( I'm just confused. I don't know what I'm talking about.
 
The general series involves the Bernouilli numbers. Are you familiar with them?.
 
The Bernouilli numbers are the numbers at the end of the expression that defines each sum of even powers.

i.e.

\(\displaystyle \sum_{k=1}^{n}k^{2}=\frac{n^{3}}{3}+\frac{n^{2}}{2}+\underbrace{\boxed{\frac{1}{6}}}_{\text{B2}}n\)

\(\displaystyle \sum_{k=1}^{n}k^{4}=\frac{n^{5}}{5}+\frac{n^{4}}{2}+\frac{n^{3}}{3}-\underbrace{\boxed{\frac{1}{30}}}_{\text{B4}}n\)

\(\displaystyle \sum_{k=1}^{n}k^{6}=\frac{1}{7}n^{7}+\frac{1}{2}n^{6}+\frac{1}{2}n^{5}-\frac{1}{6}n^{3}+\underbrace{\boxed{\frac{1}{42}}}_{\text{B6}}n\)

and on and on. The odd powers do not have them because Bernouilli numbers are 0 at the odds. The odd powers have an ending term with an n^2.

Google Benouilli numbers and sums of powers. You will find something. In order to do it any justice, it is too much for me to go into here
 
Hello,

I am finishing up with this exact portfolio, and I need a little bit of help, not with the problems though.

For the introduction, what do I have to exactly include?

Likewise, what do I have to include for the conclusion?

Thanks
 
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