Practice Problems

Find each sum.

\(\textbf{1)}\) \( \displaystyle \sum_{i=3}^{5} 3-2i \)

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The answer is \( -15 \)

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\(\,\,\,\,\,\,\displaystyle S_n=n\left(\frac{a_1+a_n}{2}\right)\)
\(\,\,\,\,\,\,\displaystyle n=\text{top } – \text{ bottom }+1 = 5-3+1=3\)
\(\,\,\,\,\,\,\displaystyle a_1=3-2(3)=3-6=-3\)
\(\,\,\,\,\,\,\displaystyle a_3=3-2(5)=3-10=-7\)
\(\,\,\,\,\,\,\displaystyle S_3=3\left(\frac{-3+-7}{2}\right)\)
\(\,\,\,\,\,\,\displaystyle S_3=3\left(\frac{-10}{2}\right)\)
\(\,\,\,\,\,\,\displaystyle S_3=3\left(-5\right)\)
\(\,\,\,\,\,\,\displaystyle S_3=-15\)

 

\(\textbf{2)}\) \( \displaystyle \sum_{i=4}^{9} 3i-5 \)

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The answer is \( 87 \)

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\(\,\,\,\,\,\,\displaystyle S_n=n\left(\frac{a_1+a_n}{2}\right)\)
\(\,\,\,\,\,\,\displaystyle n=\text{top } – \text{ bottom }+1 = 9-4+1=6\)
\(\,\,\,\,\,\,\displaystyle a_1=3(4)-5=12-5=7\)
\(\,\,\,\,\,\,\displaystyle a_6=3(9)-5=27-5=22\)
\(\,\,\,\,\,\,\displaystyle S_6=6\left(\frac{7+22}{2}\right)\)
\(\,\,\,\,\,\,\displaystyle S_6=6\left(\frac{29}{2}\right)\)
\(\,\,\,\,\,\,\displaystyle S_6=87\)

 

\(\textbf{3)}\) \( \displaystyle \sum_{i=1}^{5} 3(2)^i \)

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The answer is \( 186 \)

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\(\,\,\,\,\,\,\displaystyle S_n=a_1\left(\frac{1-r^n}{1-r}\right)\)
\(\,\,\,\,\,\,\displaystyle n=\text{top } – \text{ bottom }+1 = 5-1+1=5\)
\(\,\,\,\,\,\,\displaystyle a_1=3(2)^{(1)}=6\)
\(\,\,\,\,\,\,\displaystyle r=2\)
\(\,\,\,\,\,\,\displaystyle S_5=6\left(\frac{1-(2)^{(5)}}{1-(2)}\right)\)
\(\,\,\,\,\,\,\displaystyle S_5=6\left(\frac{1-32}{-1}\right)\)
\(\,\,\,\,\,\,\displaystyle S_5=6\left(31\right)\)
\(\,\,\,\,\,\,\displaystyle S_5=186\)

 

\(\textbf{4)}\) \( \displaystyle \sum_{i=1}^{\infty} 4\left(\frac{1}{3}\right)^{i-1} \)

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The answer is \( 6 \)

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\(\,\,\,\,\,\,S_\infty=\frac{a_1}{1-r},\,\,\,|r|\lt 1\)
\(\,\,\,\,\,\,a_1=4\left(\frac{1}{3}\right)^{1-1}=4\left(\frac{1}{3}\right)^{0}=4(1)=4\)
\(\,\,\,\,\,\,\left|\frac{1}{3}\right| \lt 1\)
\(\,\,\,\,\,\,S_\infty=\frac{4}{1-\frac{1}{3}}\)
\(\,\,\,\,\,\,S_\infty=\frac{4}{\frac{2}{3}}\)
\(\,\,\,\,\,\,S_\infty=\frac{4}{1}\cdot\frac{3}{2}\)
\(\,\,\,\,\,\,S_\infty=\frac{12}{2}\)
\(\,\,\,\,\,\,S_\infty=6\)

 

\(\textbf{5)}\) \( \displaystyle \sum_{i=2}^{5} 3i \)

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The answer is \( 42 \)

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\(\,\,\,\,\,\,\displaystyle S_n=n\left(\frac{a_1+a_n}{2}\right)\)
\(\,\,\,\,\,\,\displaystyle n=\text{top } – \text{ bottom }+1 = 5-2+1=4\)
\(\,\,\,\,\,\,\displaystyle a_1=3(2)=6\)
\(\,\,\,\,\,\,\displaystyle a_4=3(5)=15\)
\(\,\,\,\,\,\,\displaystyle S_4=4\left(\frac{6+15}{2}\right)\)
\(\,\,\,\,\,\,\displaystyle S_4=4\left(\frac{21}{2}\right)\)
\(\,\,\,\,\,\,\displaystyle S_4=42\)

 

\(\textbf{6)}\) \( \displaystyle \sum_{n=1}^{5} n^2 \)

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The answer is \( 55 \)

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\(\,\,\,\,\,\,\displaystyle \sum_{n=1}^{5} n^2\)
\(\,\,\,\,\,\,(1)^2+(2)^2+(3)^2+(4)^2+(5)^2\)
\(\,\,\,\,\,\,1+4+9+16+25=55\)

 

 

See Related Pages

\(\bullet\text{ Sigma Notation Calculator }\)
\(\,\,\,\,\,\,\,\,\text{(Symbolab.com)}\)

\(\bullet\text{ Arithmetic Sequences}\)
\(\,\,\,\,\,\,\,a_n=a_1 + d(n-1)\)

\(\bullet\text{ Geometric Sequences}\)
\(\,\,\,\,\,\,\,a_n=a_1 \cdot r^{(n-1)}…\)

\(\bullet\text{ Arithmetic Series}\)
\(\,\,\,\,\,\,\,s_n=\frac{n}{2}(a_1+a_n)…\)

\(\bullet\text{ Geometric Series}\)
\(\,\,\,\,\,\,\,s_n=a_1 \frac{1-r^n}{1-r}…\)

\(\bullet\text{ Infinite Geometric Series}\)
\(\,\,\,\,\,\,\,s_\infty = \frac{a_1}{1-r}\,\,\, |r| \lt 1…\)

\(\bullet\text{ Recursive Sequences}\)
\(\,\,\,\,\,\,\, a_{1}=2, \, a_{n+1}=a_{n}+3…\)