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Katie's question at Yahoo! Answers regarding summations

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MarkFL

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Staff member
Feb 24, 2012
13,775
Here are the questions:

Find the value of the sum of a sigma notation?

n
∑ (i+1)(2i+1)
i=1

and

n
∑(1+3i)^2
i=1

Cant figure these out, could someone show me how to do them step by step?
I have posted a link there to this thread so the OP can view my work.
 
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MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Hello Katie,

One way we could find these sums is the expand the summands, and then rely on the following formulas:

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

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

(3) \(\displaystyle \sum_{k=1}^{n}\left(k^2 \right)=\frac{n(n+1)(2n+1)}{6}\)

Also, we will use the following linearity properties of sums:

(4) \(\displaystyle \sum_{k=k_i}^{k_f}\left(a\cdot f(k) \right)=a\cdot\sum_{k=k_i}^{k_f}\left(f(k) \right)\)

(5) \(\displaystyle \sum_{k=k_i}^{k_f}\left(f(k)\pm g(k) \right)=\sum_{k=k_i}^{k_f}\left(f(k) \right)\pm\sum_{k=k_i}^{k_f}\left(g(k) \right)\)

Now, the first sum can be expanded as follows:

\(\displaystyle \sum_{k=1}^{n}\left((k+1)(2k+1) \right)=\sum_{k=1}^{n}\left(2k^2+3k+1 \right)\)

Using (5), we may write:

\(\displaystyle \sum_{k=1}^{n}\left((k+1)(2k+1) \right)=\sum_{k=1}^{n}\left(2k^2 \right)+\sum_{k=1}^{n}(3k)+\sum_{k=1}^{n}(1)\)

Using (4), we may write:

\(\displaystyle \sum_{k=1}^{n}\left((k+1)(2k+1) \right)=2\sum_{k=1}^{n}\left(k^2 \right)+3\sum_{k=1}^{n}(k)+\sum_{k=1}^{n}(1)\)

Using (1)-(3), we may write:

\(\displaystyle \sum_{k=1}^{n}\left((k+1)(2k+1) \right)=2\left(\frac{n(n+1)(2n+1)}{6} \right)+3\left(\frac{n(n+1)}{2} \right)+(n)\)

Combine terms:

\(\displaystyle \sum_{k=1}^{n}\left((k+1)(2k+1) \right)=\frac{2n(n+1)(2n+1)+9n(n+1)+6n}{6}\)

Factor numerator:

\(\displaystyle \sum_{k=1}^{n}\left((k+1)(2k+1) \right)=\frac{n\left(2(n+1)(2n+1)+9(n+1)+6 \right)}{6}\)

Expand within second factor of numerator:

\(\displaystyle \sum_{k=1}^{n}\left((k+1)(2k+1) \right)=\frac{n\left(4n^2+6n+2+9n+9+6 \right)}{6}\)

Collect like terms:

\(\displaystyle \sum_{k=1}^{n}\left((k+1)(2k+1) \right)=\frac{n\left(4n^2+15n+17 \right)}{6}\)

Now for the second sum (and following the same steps as for the first sum):

\(\displaystyle \sum_{k=1}^{n}\left((3k+1)^2 \right)=9\sum_{k=1}^{n}\left(k^2 \right)+6\sum_{k=1}^{n}(k)+\sum_{k=1}^{n}(1)\)

\(\displaystyle \sum_{k=1}^{n}\left((3k+1)^2 \right)=9\left(\frac{n(n+1)(2n+1)}{6} \right)+6\left(\frac{n(n+1)}{2} \right)+(n)\)

\(\displaystyle \sum_{k=1}^{n}\left((3k+1)^2 \right)=\frac{3n(n+1)(2n+1)+6n(n+1)+2n}{2}\)

\(\displaystyle \sum_{k=1}^{n}\left((3k+1)^2 \right)=\frac{n\left(6n^2+9n+3+6n+6+2 \right)}{2}\)

\(\displaystyle \sum_{k=1}^{n}\left((3k+1)^2 \right)=\frac{n\left(6n^2+15n+11 \right)}{2}\)

Another way we could find these sums is to treat them as linear inhomogenous difference equations:

\(\displaystyle S_{n}-S_{n-1}=an^2+bn+c\) where \(\displaystyle S_1=a+b+c\)

Now, observing that the corresponding homogeneous equation has the characteristic root $r=1$, we know the homogenous solution is:

\(\displaystyle h_n=c_1\)

And bearing in mind that none of the terms in the particular solution can be a solution to the homogenous equation, we then take for the particular solution the form:

\(\displaystyle p_n=An^3+Bn^2+Cn\)

Now we may use the method of undetermined coefficients to determing the values of the parameters $A,B,C$.

Substituting the particular solution into the difference equation, we obtain:

\(\displaystyle \left(An^3+Bn^2+Cn \right)-\left(A(n-1)^3+B(n-1)^2+C(n-1) \right)=an^2+bn+c\)

Expanding and collecting like terms, we get

\(\displaystyle 3An^2+(-3A+2B)n+(A-B+C)=an^2+bn+c\)

Equating like coefficients, we obtain the linear system:

\(\displaystyle 3A=a\implies A=\frac{a}{3}\)

\(\displaystyle -3A+2B=b\implies B=\frac{a+b}{2}\)

\(\displaystyle A-B+C=c\implies C=\frac{a+3b+6c}{6}\)

And so our particular solution is:

\(\displaystyle p_n=\frac{a}{3}n^3+\frac{a+b}{2}n^2+\frac{a+3b+6c}{6}n\)

Combining terms, we obtain:

\(\displaystyle p_n=\frac{n\left(2an^2+3(a+b)n+(a+3b+6c) \right)}{6}\)

Thus, by the principle of superposition, we find the general solution is:

\(\displaystyle S_n=h_n+p_n=c_1+\frac{n\left(2an^2+3(a+b)n+(a+3b+6c) \right)}{6}\)

Now, using the initial value, we may determine the parameter $c_1$:

\(\displaystyle S_1=c_1+\frac{2a+3(a+b)+(a+3b+6c)}{6}=a+b+c\)

\(\displaystyle c_1=a+b+c-\frac{6a+6b+6c}{6}=0\)

Thus, the general solution satisfying the given conditions is:

\(\displaystyle S_n=\frac{n\left(2an^2+3(a+b)n+(a+3b+6c) \right)}{6}\)

Now, for the first problem, we identify:

\(\displaystyle a=2,\,b=3,\,c=1\)

Hence:

\(\displaystyle S_n=\frac{n\left(4n^2+15n+17 \right)}{6}\)

And for the second problem, we identify:

\(\displaystyle a=9,\,b=6,\,c=1\)

Hence:

\(\displaystyle S_n=\frac{n\left(18n^2+45n+33 \right)}{6}=\frac{n\left(6n^2+15n+11 \right)}{2}\)