Product notation, also called pi notation, is a compact way to represent multiplying a sequence of terms. The product symbol tells you where to start, where to stop, and what expression to multiply each time. These problems practice expanding finite products, simplifying products with constants and powers, using telescoping products, and recognizing a famous infinite product example.
Notes
Product Notation
\(\displaystyle\prod_{n=1}^{k}a_n = a_1 \cdot a_2 \cdot a_3 \cdot \ldots \cdot a_{k-1} \cdot a_k\)
Practice Questions
Find each product
\(\textbf{1)}\) \( \displaystyle \prod_{i=3}^{5} (3-2i) \)
The answer is \( (-3)(-5)(-7)=-105 \)
\(\,\,\,\,\,\displaystyle \prod_{i=3}^{5} (3-2i)\)
\(\,\,\,\,\,=(3-2(3))(3-2(4))(3-2(5))\)
\(\,\,\,\,\,=(-3)(-5)(-7)\)
\(\,\,\,\,\,=-105\)
\(\textbf{2)}\) \( \displaystyle \prod_{i=3}^{5} (3i-5) \)
The answer is \( (4)(7)(10)=280 \)
\(\,\,\,\,\,\displaystyle \prod_{i=3}^{5} (3i-5)\)
\(\,\,\,\,\,=(3(3)-5)(3(4)-5)(3(5)-5)\)
\(\,\,\,\,\,=(4)(7)(10)\)
\(\,\,\,\,\,=280\)
\(\textbf{3)}\) \( \displaystyle \prod_{i=1}^{4} (2)^i \)
The answer is \( (2)(4)(8)(16)=1024 \)
\(\,\,\,\,\,\displaystyle \prod_{i=1}^{4} 2^i\)
\(\,\,\,\,\,=2^1\cdot2^2\cdot2^3\cdot2^4\)
\(\,\,\,\,\,=2^{1+2+3+4}\)
\(\,\,\,\,\,=2^{10}\)
\(\,\,\,\,\,=1024\)
\(\textbf{4)}\) \( \displaystyle \prod_{i=1}^{4} i \)
The answer is \(24\)
\(\,\,\,\,\,\displaystyle \prod_{i=1}^{4} i\)
\(\,\,\,\,\,=(1)(2)(3)(4)\)
\(\,\,\,\,\,=24\)
\(\textbf{5)}\) \( \displaystyle \prod_{i=1}^{5} (i+1) \)
The answer is \(720\)
\(\,\,\,\,\,\displaystyle \prod_{i=1}^{5} (i+1)\)
\(\,\,\,\,\,=(2)(3)(4)(5)(6)\)
\(\,\,\,\,\,=720\)
\(\textbf{6)}\) \( \displaystyle \prod_{i=2}^{5} 2i \)
The answer is \(1920\)
\(\,\,\,\,\,\displaystyle \prod_{i=2}^{5} 2i\)
\(\,\,\,\,\,=(2(2))(2(3))(2(4))(2(5))\)
\(\,\,\,\,\,=(4)(6)(8)(10)\)
\(\,\,\,\,\,=1920\)
\(\textbf{7)}\) \( \displaystyle \prod_{i=1}^{3} (5-i) \)
The answer is \(24\)
\(\,\,\,\,\,\displaystyle \prod_{i=1}^{3} (5-i)\)
\(\,\,\,\,\,=(5-1)(5-2)(5-3)\)
\(\,\,\,\,\,=(4)(3)(2)\)
\(\,\,\,\,\,=24\)
\(\textbf{8)}\) \( \displaystyle \prod_{i=1}^{4} (2i+1) \)
The answer is \(945\)
\(\,\,\,\,\,\displaystyle \prod_{i=1}^{4} (2i+1)\)
\(\,\,\,\,\,=(3)(5)(7)(9)\)
\(\,\,\,\,\,=945\)
\(\textbf{9)}\) \( \displaystyle \prod_{i=0}^{3} (i+2) \)
The answer is \(120\)
\(\,\,\,\,\,\displaystyle \prod_{i=0}^{3} (i+2)\)
\(\,\,\,\,\,=(2)(3)(4)(5)\)
\(\,\,\,\,\,=120\)
\(\textbf{10)}\) \( \displaystyle \prod_{i=1}^{4} 3 \)
The answer is \(81\)
\(\,\,\,\,\,\displaystyle \prod_{i=1}^{4} 3\)
\(\,\,\,\,\,=(3)(3)(3)(3)\)
\(\,\,\,\,\,=81\)
\(\textbf{11)}\) \( \displaystyle \prod_{i=2}^{4} i^2 \)
The answer is \(576\)
\(\,\,\,\,\,\displaystyle \prod_{i=2}^{4} i^2\)
\(\,\,\,\,\,=(2^2)(3^2)(4^2)\)
\(\,\,\,\,\,=(4)(9)(16)\)
\(\,\,\,\,\,=576\)
\(\textbf{12)}\) \( \displaystyle \prod_{i=1}^{3} (4i-1) \)
The answer is \(231\)
\(\,\,\,\,\,\displaystyle \prod_{i=1}^{3} (4i-1)\)
\(\,\,\,\,\,=(3)(7)(11)\)
\(\,\,\,\,\,=231\)
\(\textbf{13)}\) \( \displaystyle \prod_{i=1}^{4} \frac{i+1}{i} \)
The answer is \(5\)
\(\,\,\,\,\,\displaystyle \prod_{i=1}^{4} \frac{i+1}{i}\)
\(\,\,\,\,\,=\frac{2}{1}\cdot\frac{3}{2}\cdot\frac{4}{3}\cdot\frac{5}{4}\)
\(\,\,\,\,\,=5\)
\(\textbf{14)}\) \( \displaystyle \prod_{i=1}^{3} (-i) \)
The answer is \(-6\)
\(\,\,\,\,\,\displaystyle \prod_{i=1}^{3} (-i)\)
\(\,\,\,\,\,=(-1)(-2)(-3)\)
\(\,\,\,\,\,=-6\)
\(\textbf{15)}\) \( \displaystyle \prod_{i=2}^{5} (i-1) \)
The answer is \(24\)
\(\,\,\,\,\,\displaystyle \prod_{i=2}^{5} (i-1)\)
\(\,\,\,\,\,=(1)(2)(3)(4)\)
\(\,\,\,\,\,=24\)
\(\textbf{16)}\) \( \displaystyle \prod_{i=1}^{5} \frac{i}{i+1} \)
The answer is \(\frac{1}{6}\)
\(\,\,\,\,\,\displaystyle \prod_{i=1}^{5} \frac{i}{i+1}\)
\(\,\,\,\,\,=\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot\frac{4}{5}\cdot\frac{5}{6}\)
\(\,\,\,\,\,=\frac{1}{6}\)
\(\textbf{17)}\) \( \displaystyle \prod_{i=1}^{4} (2^i-1) \)
The answer is \(315\)
\(\,\,\,\,\,\displaystyle \prod_{i=1}^{4} (2^i-1)\)
\(\,\,\,\,\,=(2^1-1)(2^2-1)(2^3-1)(2^4-1)\)
\(\,\,\,\,\,=(1)(3)(7)(15)\)
\(\,\,\,\,\,=315\)
\(\textbf{18)}\) \( \displaystyle \prod_{i=1}^{4} \frac{2i}{i+1} \)
The answer is \(\frac{16}{5}\)
\(\,\,\,\,\,\displaystyle \prod_{i=1}^{4} \frac{2i}{i+1}\)
\(\,\,\,\,\,=\frac{2}{2}\cdot\frac{4}{3}\cdot\frac{6}{4}\cdot\frac{8}{5}\)
\(\,\,\,\,\,=1\cdot\frac{4}{3}\cdot\frac{3}{2}\cdot\frac{8}{5}\)
\(\,\,\,\,\,=\frac{16}{5}\)
\(\textbf{19)}\) Wallis’ product says \(\displaystyle \frac{\pi}{2}=\prod_{n=1}^{\infty}\frac{(2n)(2n)}{(2n-1)(2n+1)}\). Find the first three-factor approximation.
The answer is \(\frac{256}{175}\).
\(\,\,\,\,\,\displaystyle \prod_{n=1}^{3}\frac{(2n)(2n)}{(2n-1)(2n+1)}\)
\(\,\,\,\,\,=\frac{(2)(2)}{(1)(3)}\cdot\frac{(4)(4)}{(3)(5)}\cdot\frac{(6)(6)}{(5)(7)}\)
\(\,\,\,\,\,=\frac{4}{3}\cdot\frac{16}{15}\cdot\frac{36}{35}\)
\(\,\,\,\,\,=\frac{2304}{1575}\)
\(\,\,\,\,\,=\frac{256}{175}\)
\(\textbf{20)}\)\(\displaystyle \prod_{n=2}^{\infty}\frac{n^3-1}{n^3+1}\)
The answer is \(\frac{2}{3}\).
\(\,\,\,\,\,\displaystyle \prod_{n=2}^{\infty}\frac{n^3-1}{n^3+1}\)
\(\,\,\,\,\,\displaystyle n^3-1=(n-1)(n^2+n+1)\)
\(\,\,\,\,\,\displaystyle n^3+1=(n+1)(n^2-n+1)\)
\(\,\,\,\,\,\displaystyle \frac{n^3-1}{n^3+1}=\frac{(n-1)(n^2+n+1)}{(n+1)(n^2-n+1)}\)
\(\,\,\,\,\,\displaystyle n^2+n+1=(n+1)^2-(n+1)+1\)
\(\,\,\,\,\,\displaystyle \prod_{n=2}^{N}\frac{n^3-1}{n^3+1}=\prod_{n=2}^{N}\frac{n-1}{n+1}\cdot\prod_{n=2}^{N}\frac{n^2+n+1}{n^2-n+1}\)
\(\,\,\,\,\,\displaystyle \prod_{n=2}^{N}\frac{n-1}{n+1}=\frac{1\cdot2\cdot3\cdots(N-1)}{3\cdot4\cdot5\cdots(N+1)}=\frac{2}{N(N+1)}\)
\(\,\,\,\,\,\displaystyle \prod_{n=2}^{N}\frac{n^2+n+1}{n^2-n+1}=\frac{N^2+N+1}{3}\)
\(\,\,\,\,\,\displaystyle \prod_{n=2}^{N}\frac{n^3-1}{n^3+1}=\frac{2}{N(N+1)}\cdot\frac{N^2+N+1}{3}\)
\(\,\,\,\,\,\displaystyle \prod_{n=2}^{N}\frac{n^3-1}{n^3+1}=\frac{2(N^2+N+1)}{3N(N+1)}\)
\(\,\,\,\,\,\displaystyle \prod_{n=2}^{\infty}\frac{n^3-1}{n^3+1}=\lim_{N\to\infty}\frac{2(N^2+N+1)}{3N(N+1)}\)
\(\,\,\,\,\,\displaystyle =\frac{2}{3}\)
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