MHB [Series] State the first four terms and find the nth term

[bunyonb]

What is the procedure for finding the unknown term(end value in this scenario) in a series? For example

$$

\sum_{r=1}^{n}{2r+3}
$$

My Attempt was to simply state the first four terms and then simply add the nth term as it is:

2(1)+3=5
2(2)+3=7
2(3)+3=9
2(4)+3=11
2(n)+3=2n+3


Total=5+7+9+11+2n+3=35+2n

Would this be a correct procedure or is here something I am misunderstanding? I cannot remember if you are supposed to multiply the last value with the sequence or not.

 

[MarkFL]

Your statement of the first 4 terms, and the $n$th term are correct, however, if you wish to find the sum, I would proceed as follows:

$$S=\sum_{r=1}^{n}(2r+3)=2\sum_{r=1}^{n}(r)+3\sum_{r=1}^{n}(1)=2\left(\frac{n(n+1)}{2}\right)+3(n)=n(n+1)+3n=n^2+n+3n=n^2+4n=n(n+4)$$

 

[bunyonb]

Your statement of the first 4 terms, and the $n$th term are correct, however, if you wish to find the sum, I would proceed as follows:

$$S=\sum_{r=1}^{n}(2r+3)=2\sum_{r=1}^{n}(r)+3\sum_{r=1}^{n}(1)=2\left(\frac{n(n+1)}{2}\right)+3(n)=n(n+1)+3n=n^2+n+3n=n^2+4n=n(n+4)$$

So the procedure to find the nth term is to solve for n algebraically?

 

[MarkFL]

So the procedure to find the nth term is to solve for n algebraically?

No, you correctly found the $n$th term, but if we wish to actually sum the series, then we have to apply some summation techniques. :D

 

[bunyonb]

summation techniques. :D

Well there you have it. I do not know summation techniques. That's what I need to study.. Thanks. It is much easier when i understand the correct terminologies and terms so i can have easier means to reference or look it up. Half of my problems in mathematics is not knowing what to look for because i don't know what something or a procedure is called.

 

[MarkFL]

We could also derive the closed-forum for the sum directly by stating it in the following linear inhomogeneous difference equation (recursion):

$$S_{n}-S_{n-1}=2n+3$$ where $S_1=5$

Instead of relying on memorized summation formulas.

Now, we see the characteristic equation has the root $r=1$, and so the homogeneous solution is:

$$h_n=c_1$$

Now, observing that the RHS of the difference equation has a constant term, and noting that the homogeneous solution is itself a constant, we must then assume the particular solution will take the form:

$$p_n=n(An+B)=An^2+Bn$$

Now, we can use the method of undetermined coefficients to find $A$ and $B$...so we substitute the particular solution into the difference equation:

$$\left(An^2+Bn\right)-\left(A(n-1)^2+B(n-1)\right)=2n+3$$

$$An^2+Bn-A(n^2-2n+1)-B(n-1)=2n+3$$

$$An^2+Bn-An^2+2An-A-Bn+B=2n+3$$

$$2An+(-A+B)=2n+3$$

Equating coefficients, we obtain:

$$2A=2\implies A=1$$

$$-A+B=3\implies B=4$$

And so our particular solution is:

$$p_n=n^2+4n$$

And by the principle of superposition, the general solution is:

$$S_n=h_n+p_n=c_1+n^2+4n$$

Now, we use the initial condition to determine the parameter:

$$S_1=c_1+1^2+4(1)=c_1+5=5\implies c_1=0$$

And so the solution satisfying all conditions is:

$$S_n=n^2+4n=n(n+4)$$