How do you put 5 + 16 + 33 + ... + 3n^2 + 2n into a formula?

Question

I do not understand how to make it into a formula, an explanation along with the answer would be perfect!

It can't be in series notation. Must be in a formula so that I can add it to another formula.

Answer 1

Write down a few more terms in your series :
5 + 16 + 33 + 56 + 85 + 120 + ... + 3n^2 + 2n

As you sum these, your sequence will be :
5, 21, 54, 110, 195, 315, ...

Now take the difference between terms,
e.g. 21 - 5 = 16, 54 - 21 = 33, etc.
This gives you the first difference sequence :
16, 33, 56, 85, 120, ...

Now take the difference between these latter terms,
e.g. 33 - 16 = 17, etc.
This gives you the second difference sequence :
17, 23, 29, 35, ...

And again, keep taking the difference until the
numbers become equal. Of course, this will not
work for every series, but it does work for this
nicely behaving polynomial series.
So the third difference sequence is :
6, 6, 6, ...

The numbers are now equal and we can stop at
this so-called, third-order sequence.

In general, a sequence of numbers whose k(th) difference
is constant, is called a sequence of the k(th)-order, and
the numbers are related by the polynomial formula :
N = am^k + bm^(k-1) + cm^(k-2) + ... + xm^2 + ym + z
where a,b,c,...,x,y,z are the equation coefficients
to be calculated, k is the order of the sequence, and m
takes on the values of the natural numbers 1,2,3,4,...
I used m so as not confuse with your n.

Your equation will be third-order, so k = 3, so the
equation is: N = am^3 + bm^2 + cm + d

Now the original sequence was 5, 21, 54, 110, 195, 315, ...
So when m = 1, N will be 5, when m = 2, N = 21, etc.

So now we set up these simultaneous equations
and solve for a, b and c and d.
We have then:

For m = 1, 5 = a + b + c + d
For m = 2, 21 = 8a + 4b + 2c + d
For m = 3, 54 = 27a + 9b + 3c + d
For m = 4, 110 = 64a + 16b + 4c + d

This is enough to establish the final equation -
4 equations for 4 variables.

After you've done all the hard work, you should find that:
a = 1, b = 5/2, c = 3/2 and d = 0.

Substituting these back into the third-order equation
gives you the formula:
N = m^3 + (5/2)m^2 + (3/2)m

On simplifying:
N = m(2m^2 + 5m + 3) / 2

On factorising:
N = m(m + 1)(2m + 3) / 2
This is your final formula.

Now when you substitute m = 1, N = 5.
m = 2 gives N = 21, m = 3 gives N = 54, etc.,
which is your sum to m terms.

Answer 2

WE are already given the nth term as 3n^2+2n.
So,
5+16+33+.......3n^2+2n=Sum(3n^2+2n)
=3*Summation(n^2)+2*Sum(n)
Sum of frst n squares of natural numbers is given by n(n+1)(2n+1)/6 and the sum of first n natural numbers is n(n+1)/2.

So, we get
3n(n+1)(2n+1)/6+2n(n+1)/2
=n(n+1)(2n+1)/2+n(n+1)
=[n(n+1)/2][2n+1+2]
=n(n+1)(2n+3)/2.

You can verify that this is the correct answer

Answer 3

they already in fact gave you the formula which is the last n interval in the series.

the series begins at i=1 where i is constantly adding +1 for next numeral, i then i+1 then i+2 for third number,ect. until the last numeral which is n and the function is equal to,
f(x)=3i^2+2i

this is why the first number is 3(1)^2+2(1)=5

i=1+1=2 for 2nd numeral

likewise the next number is 3(2)^2+2(2)=16

i=1+2=3 for third numeral

likewise the next numeral is 3(3)^2+3(3)=33

until final numeral which is i=n so sub n for i

final numeral is 3(n)^2+2(n)=3n^2+2n
--------------------------------------------------------------

formula is going to be using standard i notation in the series

3Ei^2+2Ei

=3[n(n+1)(2n+1)/6]+2[n(n+1)/2]

=(n(n+1)(2n+1)/2)+(2n(n+1)/2)

=([n(n+1)(2n+1) + 2n(n+1)]/2) this is the formula and I believe you can switch the n variable for x or any other variable to add to another equation

Answer 4

here
Sn = 5 + 16 + 33 + ................+ (3*n^2 + 2*n)

Its nth term is 3n^2 + 2n
so
Sn = Summation of (3n^2 + 2n)
where n = 1 to n

Answer 5

It is just the summation of the equation 3n^2 + 2n.
∑(3n^2 + 2n)