Summation notation, also called sigma notation, is a compact way to write repeated addition. These problems include finite arithmetic sums, finite geometric sums, infinite geometric sums, and direct expansion of sums. The main idea is to carefully substitute each integer value from the lower limit to the upper limit, then add the resulting terms.
Practice Problems
Find each sum.
\(\textbf{1)}\) \( \displaystyle \sum_{i=3}^{5} 3-2i \)
The answer is \( -15 \)
\(\,\,\,\,\,\,\displaystyle S_n=n\left(\frac{a_1+a_n}{2}\right)\)
\(\,\,\,\,\,\,\displaystyle n=\text{top } – \text{ bottom }+1 = 5-3+1=3\)
\(\,\,\,\,\,\,\displaystyle a_1=3-2(3)=3-6=-3\)
\(\,\,\,\,\,\,\displaystyle a_3=3-2(5)=3-10=-7\)
\(\,\,\,\,\,\,\displaystyle S_3=3\left(\frac{-3+-7}{2}\right)\)
\(\,\,\,\,\,\,\displaystyle S_3=3\left(\frac{-10}{2}\right)\)
\(\,\,\,\,\,\,\displaystyle S_3=3\left(-5\right)\)
\(\,\,\,\,\,\,\displaystyle S_3=-15\)
\(\textbf{2)}\) \( \displaystyle \sum_{i=4}^{9} 3i-5 \)
The answer is \( 87 \)
\(\,\,\,\,\,\,\displaystyle S_n=n\left(\frac{a_1+a_n}{2}\right)\)
\(\,\,\,\,\,\,\displaystyle n=\text{top } – \text{ bottom }+1 = 9-4+1=6\)
\(\,\,\,\,\,\,\displaystyle a_1=3(4)-5=12-5=7\)
\(\,\,\,\,\,\,\displaystyle a_6=3(9)-5=27-5=22\)
\(\,\,\,\,\,\,\displaystyle S_6=6\left(\frac{7+22}{2}\right)\)
\(\,\,\,\,\,\,\displaystyle S_6=6\left(\frac{29}{2}\right)\)
\(\,\,\,\,\,\,\displaystyle S_6=87\)
\(\textbf{3)}\) \( \displaystyle \sum_{i=1}^{5} 3(2)^i \)
The answer is \( 186 \)
\(\,\,\,\,\,\,\displaystyle S_n=a_1\left(\frac{1-r^n}{1-r}\right)\)
\(\,\,\,\,\,\,\displaystyle n=\text{top } – \text{ bottom }+1 = 5-1+1=5\)
\(\,\,\,\,\,\,\displaystyle a_1=3(2)^{(1)}=6\)
\(\,\,\,\,\,\,\displaystyle r=2\)
\(\,\,\,\,\,\,\displaystyle S_5=6\left(\frac{1-(2)^{(5)}}{1-(2)}\right)\)
\(\,\,\,\,\,\,\displaystyle S_5=6\left(\frac{1-32}{-1}\right)\)
\(\,\,\,\,\,\,\displaystyle S_5=6\left(31\right)\)
\(\,\,\,\,\,\,\displaystyle S_5=186\)
\(\textbf{4)}\) \( \displaystyle \sum_{i=1}^{\infty} 4\left(\frac{1}{3}\right)^{i-1} \)
The answer is \( 6 \)
\(\,\,\,\,\,\,S_\infty=\frac{a_1}{1-r},\,\,\,|r|\lt 1\)
\(\,\,\,\,\,\,a_1=4\left(\frac{1}{3}\right)^{1-1}=4\left(\frac{1}{3}\right)^{0}=4(1)=4\)
\(\,\,\,\,\,\,\left|\frac{1}{3}\right| \lt 1\)
\(\,\,\,\,\,\,S_\infty=\frac{4}{1-\frac{1}{3}}\)
\(\,\,\,\,\,\,S_\infty=\frac{4}{\frac{2}{3}}\)
\(\,\,\,\,\,\,S_\infty=\frac{4}{1}\cdot\frac{3}{2}\)
\(\,\,\,\,\,\,S_\infty=\frac{12}{2}\)
\(\,\,\,\,\,\,S_\infty=6\)
\(\textbf{5)}\) \( \displaystyle \sum_{i=2}^{5} 3i \)
The answer is \( 42 \)
\(\,\,\,\,\,\,\displaystyle S_n=n\left(\frac{a_1+a_n}{2}\right)\)
\(\,\,\,\,\,\,\displaystyle n=\text{top } – \text{ bottom }+1 = 5-2+1=4\)
\(\,\,\,\,\,\,\displaystyle a_1=3(2)=6\)
\(\,\,\,\,\,\,\displaystyle a_4=3(5)=15\)
\(\,\,\,\,\,\,\displaystyle S_4=4\left(\frac{6+15}{2}\right)\)
\(\,\,\,\,\,\,\displaystyle S_4=4\left(\frac{21}{2}\right)\)
\(\,\,\,\,\,\,\displaystyle S_4=42\)
\(\textbf{6)}\) \( \displaystyle \sum_{n=1}^{5} n^2 \)
The answer is \( 55 \)
\(\,\,\,\,\,\,\displaystyle \sum_{n=1}^{5} n^2\)
\(\,\,\,\,\,\,(1)^2+(2)^2+(3)^2+(4)^2+(5)^2\)
\(\,\,\,\,\,\,1+4+9+16+25=55\)
\(\textbf{7)}\) \( \displaystyle \sum_{k=1}^{6} (2k+1) \)
The answer is \( 48 \)
\(\,\,\,\,\,\,\displaystyle \sum_{k=1}^{6} (2k+1)\)
\(\,\,\,\,\,\,(2(1)+1)+(2(2)+1)+(2(3)+1)+(2(4)+1)+(2(5)+1)+(2(6)+1)\)
\(\,\,\,\,\,\,3+5+7+9+11+13\)
\(\,\,\,\,\,\,48\)
\(\textbf{8)}\) \( \displaystyle \sum_{j=0}^{4} 5(3)^j \)
The answer is \( 605 \)
\(\,\,\,\,\,\,\displaystyle S_n=a_1\left(\frac{1-r^n}{1-r}\right)\)
\(\,\,\,\,\,\,\displaystyle n=4-0+1=5\)
\(\,\,\,\,\,\,\displaystyle a_1=5(3)^0=5\)
\(\,\,\,\,\,\,\displaystyle r=3\)
\(\,\,\,\,\,\,\displaystyle S_5=5\left(\frac{1-3^5}{1-3}\right)\)
\(\,\,\,\,\,\,\displaystyle S_5=5\left(\frac{1-243}{-2}\right)\)
\(\,\,\,\,\,\,\displaystyle S_5=5(121)\)
\(\,\,\,\,\,\,\displaystyle S_5=605\)
\(\textbf{9)}\) \( \displaystyle \sum_{n=2}^{6} (n^2-1) \)
The answer is \( 85 \)
\(\,\,\,\,\,\,\displaystyle \sum_{n=2}^{6} (n^2-1)\)
\(\,\,\,\,\,\,\left(2^2-1\right)+\left(3^2-1\right)+\left(4^2-1\right)+\left(5^2-1\right)+\left(6^2-1\right)\)
\(\,\,\,\,\,\,3+8+15+24+35\)
\(\,\,\,\,\,\,85\)
\(\textbf{10)}\) \( \displaystyle \sum_{i=1}^{4} (4i-7) \)
The answer is \( 12 \)
\(\,\,\,\,\,\,\displaystyle \sum_{i=1}^{4} (4i-7)\)
\(\,\,\,\,\,\,(4(1)-7)+(4(2)-7)+(4(3)-7)+(4(4)-7)\)
\(\,\,\,\,\,\,-3+1+5+9\)
\(\,\,\,\,\,\,12\)
\(\textbf{11)}\) \( \displaystyle \sum_{m=1}^{\infty} 6\left(\frac{1}{2}\right)^{m-1} \)
The answer is \( 12 \)
\(\,\,\,\,\,\,S_\infty=\frac{a_1}{1-r},\,\,\,|r|\lt 1\)
\(\,\,\,\,\,\,a_1=6\left(\frac{1}{2}\right)^{1-1}=6\)
\(\,\,\,\,\,\,r=\frac{1}{2}\)
\(\,\,\,\,\,\,\left|\frac{1}{2}\right|\lt 1\)
\(\,\,\,\,\,\,S_\infty=\frac{6}{1-\frac{1}{2}}\)
\(\,\,\,\,\,\,S_\infty=\frac{6}{\frac{1}{2}}\)
\(\,\,\,\,\,\,S_\infty=12\)
\(\textbf{12)}\) \( \displaystyle \sum_{x=3}^{7} (10-x) \)
The answer is \( 25 \)
\(\,\,\,\,\,\,\displaystyle \sum_{x=3}^{7} (10-x)\)
\(\,\,\,\,\,\,(10-3)+(10-4)+(10-5)+(10-6)+(10-7)\)
\(\,\,\,\,\,\,7+6+5+4+3\)
\(\,\,\,\,\,\,25\)
\(\textbf{13)}\) \( \displaystyle \sum_{i=1}^{5} \frac{i}{2} \)
The answer is \( \frac{15}{2} \)
\(\,\,\,\,\,\,\displaystyle \sum_{i=1}^{5} \frac{i}{2}\)
\(\,\,\,\,\,\,\frac{1}{2}+\frac{2}{2}+\frac{3}{2}+\frac{4}{2}+\frac{5}{2}\)
\(\,\,\,\,\,\,\frac{1+2+3+4+5}{2}\)
\(\,\,\,\,\,\,\frac{15}{2}\)
\(\textbf{14)}\) \( \displaystyle \sum_{i=0}^{3} (2^i+1) \)
The answer is \( 19 \)
\(\,\,\,\,\,\,\displaystyle \sum_{i=0}^{3} (2^i+1)\)
\(\,\,\,\,\,\,(2^0+1)+(2^1+1)+(2^2+1)+(2^3+1)\)
\(\,\,\,\,\,\,2+3+5+9\)
\(\,\,\,\,\,\,19\)
\(\textbf{15)}\) \( \displaystyle \sum_{r=1}^{4} 7 \)
The answer is \( 28 \)
\(\,\,\,\,\,\,\displaystyle \sum_{r=1}^{4} 7\)
\(\,\,\,\,\,\,7+7+7+7\)
\(\,\,\,\,\,\,28\)
\(\textbf{16)}\) \( \displaystyle \sum_{p=1}^{4} (p^3) \)
The answer is \( 100 \)
\(\,\,\,\,\,\,\displaystyle \sum_{p=1}^{4} p^3\)
\(\,\,\,\,\,\,(1)^3+(2)^3+(3)^3+(4)^3\)
\(\,\,\,\,\,\,1+8+27+64\)
\(\,\,\,\,\,\,100\)
\(\textbf{17)}\) \( \displaystyle \sum_{i=2}^{6} (5i+2) \)
The answer is \( 110 \)
\(\,\,\,\,\,\,\displaystyle S_n=n\left(\frac{a_1+a_n}{2}\right)\)
\(\,\,\,\,\,\,\displaystyle n=6-2+1=5\)
\(\,\,\,\,\,\,\displaystyle a_1=5(2)+2=12\)
\(\,\,\,\,\,\,\displaystyle a_5=5(6)+2=32\)
\(\,\,\,\,\,\,\displaystyle S_5=5\left(\frac{12+32}{2}\right)\)
\(\,\,\,\,\,\,\displaystyle S_5=5(22)\)
\(\,\,\,\,\,\,\displaystyle S_5=110\)
\(\textbf{18)}\) \( \displaystyle \sum_{n=1}^{4} 2(3)^{n-1} \)
The answer is \( 80 \)
\(\,\,\,\,\,\,\displaystyle S_n=a_1\left(\frac{1-r^n}{1-r}\right)\)
\(\,\,\,\,\,\,\displaystyle n=4\)
\(\,\,\,\,\,\,\displaystyle a_1=2(3)^{1-1}=2\)
\(\,\,\,\,\,\,\displaystyle r=3\)
\(\,\,\,\,\,\,\displaystyle S_4=2\left(\frac{1-3^4}{1-3}\right)\)
\(\,\,\,\,\,\,\displaystyle S_4=2\left(\frac{1-81}{-2}\right)\)
\(\,\,\,\,\,\,\displaystyle S_4=2(40)\)
\(\,\,\,\,\,\,\displaystyle S_4=80\)
\(\textbf{19)}\) \( \displaystyle \sum_{i=1}^{\infty} 10\left(-\frac{1}{5}\right)^{i-1} \)
The answer is \( \frac{25}{3} \)
\(\,\,\,\,\,\,S_\infty=\frac{a_1}{1-r},\,\,\,|r|\lt 1\)
\(\,\,\,\,\,\,a_1=10\left(-\frac{1}{5}\right)^{1-1}=10\)
\(\,\,\,\,\,\,r=-\frac{1}{5}\)
\(\,\,\,\,\,\,\left|-\frac{1}{5}\right|\lt 1\)
\(\,\,\,\,\,\,S_\infty=\frac{10}{1-\left(-\frac{1}{5}\right)}\)
\(\,\,\,\,\,\,S_\infty=\frac{10}{\frac{6}{5}}\)
\(\,\,\,\,\,\,S_\infty=10\cdot\frac{5}{6}\)
\(\,\,\,\,\,\,S_\infty=\frac{25}{3}\)
\(\textbf{20)}\) \( \displaystyle \sum_{q=1}^{5} (3q^2-2q) \)
The answer is \( 135 \)
\(\,\,\,\,\,\,\displaystyle \sum_{q=1}^{5} (3q^2-2q)\)
\(\,\,\,\,\,\,\left(3(1)^2-2(1)\right)+\left(3(2)^2-2(2)\right)+\left(3(3)^2-2(3)\right)+\left(3(4)^2-2(4)\right)+\left(3(5)^2-2(5)\right)\)
\(\,\,\,\,\,\,1+8+21+40+65\)
\(\,\,\,\,\,\,135\)
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