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 \( (-3)(-5)(-7)=-105 \)
\(\,\,\,\,\,\displaystyle \prod_{i=3}^{5} (3-2i)\)
\(\,\,\,\,\,=(3-2(3))(3-2(4))(3-2(5))\)
\(\,\,\,\,\,=(-3)(-5)(-7)\)
\(\,\,\,\,\,=-105\)
\(\,\,\,\,\,\displaystyle \prod_{i=3}^{5} (3-2i)\)
\(\,\,\,\,\,=(3-2(3))(3-2(4))(3-2(5))\)
\(\,\,\,\,\,=(-3)(-5)(-7)\)
\(\,\,\,\,\,=-105\)
\(\textbf{2)}\) \( \displaystyle \prod_{i=3}^{5} (3i-5) \)
The answer is \( (4)(7)(10)=280 \)
\(\,\,\,\,\,\displaystyle \prod_{i=3}^{5} (3i-5)\)
\(\,\,\,\,\,=(3(3)-5)(3(4)-5)(3(5)-5)\)
\(\,\,\,\,\,=(4)(7)(10)\)
\(\,\,\,\,\,=280\)
\(\,\,\,\,\,\displaystyle \prod_{i=3}^{5} (3i-5)\)
\(\,\,\,\,\,=(3(3)-5)(3(4)-5)(3(5)-5)\)
\(\,\,\,\,\,=(4)(7)(10)\)
\(\,\,\,\,\,=280\)
\(\textbf{3)}\) \( \displaystyle \prod_{i=1}^{4} (2)^i \)
The answer is \( (2)(4)(8)(16)=1024 \)
\(\,\,\,\,\,\displaystyle \prod_{i=1}^{4} 2^i\)
\(\,\,\,\,\,=2^1\cdot2^2\cdot2^3\cdot2^4\)
\(\,\,\,\,\,=2^{1+2+3+4}\)
\(\,\,\,\,\,=2^{10}\)
\(\,\,\,\,\,=1024\)
\(\,\,\,\,\,\displaystyle \prod_{i=1}^{4} 2^i\)
\(\,\,\,\,\,=2^1\cdot2^2\cdot2^3\cdot2^4\)
\(\,\,\,\,\,=2^{1+2+3+4}\)
\(\,\,\,\,\,=2^{10}\)
\(\,\,\,\,\,=1024\)
\(\textbf{4)}\) \( \displaystyle \prod_{i=1}^{4} i \)
The answer is \(24\)
\(\,\,\,\,\,\displaystyle \prod_{i=1}^{4} i\)
\(\,\,\,\,\,=(1)(2)(3)(4)\)
\(\,\,\,\,\,=24\)
\(\,\,\,\,\,\displaystyle \prod_{i=1}^{4} i\)
\(\,\,\,\,\,=(1)(2)(3)(4)\)
\(\,\,\,\,\,=24\)
\(\textbf{5)}\) \( \displaystyle \prod_{i=1}^{5} (i+1) \)
The answer is \(720\)
\(\,\,\,\,\,\displaystyle \prod_{i=1}^{5} (i+1)\)
\(\,\,\,\,\,=(2)(3)(4)(5)(6)\)
\(\,\,\,\,\,=720\)
\(\,\,\,\,\,\displaystyle \prod_{i=1}^{5} (i+1)\)
\(\,\,\,\,\,=(2)(3)(4)(5)(6)\)
\(\,\,\,\,\,=720\)
\(\textbf{6)}\) \( \displaystyle \prod_{i=2}^{5} 2i \)
The answer is \(1920\)
\(\,\,\,\,\,\displaystyle \prod_{i=2}^{5} 2i\)
\(\,\,\,\,\,=(2(2))(2(3))(2(4))(2(5))\)
\(\,\,\,\,\,=(4)(6)(8)(10)\)
\(\,\,\,\,\,=1920\)
\(\,\,\,\,\,\displaystyle \prod_{i=2}^{5} 2i\)
\(\,\,\,\,\,=(2(2))(2(3))(2(4))(2(5))\)
\(\,\,\,\,\,=(4)(6)(8)(10)\)
\(\,\,\,\,\,=1920\)
\(\textbf{7)}\) \( \displaystyle \prod_{i=1}^{3} (5-i) \)
The answer is \(24\)
\(\,\,\,\,\,\displaystyle \prod_{i=1}^{3} (5-i)\)
\(\,\,\,\,\,=(5-1)(5-2)(5-3)\)
\(\,\,\,\,\,=(4)(3)(2)\)
\(\,\,\,\,\,=24\)
\(\,\,\,\,\,\displaystyle \prod_{i=1}^{3} (5-i)\)
\(\,\,\,\,\,=(5-1)(5-2)(5-3)\)
\(\,\,\,\,\,=(4)(3)(2)\)
\(\,\,\,\,\,=24\)
\(\textbf{8)}\) \( \displaystyle \prod_{i=1}^{4} (2i+1) \)
The answer is \(945\)
\(\,\,\,\,\,\displaystyle \prod_{i=1}^{4} (2i+1)\)
\(\,\,\,\,\,=(3)(5)(7)(9)\)
\(\,\,\,\,\,=945\)
\(\,\,\,\,\,\displaystyle \prod_{i=1}^{4} (2i+1)\)
\(\,\,\,\,\,=(3)(5)(7)(9)\)
\(\,\,\,\,\,=945\)
\(\textbf{9)}\) \( \displaystyle \prod_{i=0}^{3} (i+2) \)
The answer is \(120\)
\(\,\,\,\,\,\displaystyle \prod_{i=0}^{3} (i+2)\)
\(\,\,\,\,\,=(2)(3)(4)(5)\)
\(\,\,\,\,\,=120\)
\(\,\,\,\,\,\displaystyle \prod_{i=0}^{3} (i+2)\)
\(\,\,\,\,\,=(2)(3)(4)(5)\)
\(\,\,\,\,\,=120\)
\(\textbf{10)}\) \( \displaystyle \prod_{i=1}^{4} 3 \)
The answer is \(81\)
\(\,\,\,\,\,\displaystyle \prod_{i=1}^{4} 3\)
\(\,\,\,\,\,=(3)(3)(3)(3)\)
\(\,\,\,\,\,=81\)
\(\,\,\,\,\,\displaystyle \prod_{i=1}^{4} 3\)
\(\,\,\,\,\,=(3)(3)(3)(3)\)
\(\,\,\,\,\,=81\)
\(\textbf{11)}\) \( \displaystyle \prod_{i=2}^{4} i^2 \)
The answer is \(576\)
\(\,\,\,\,\,\displaystyle \prod_{i=2}^{4} i^2\)
\(\,\,\,\,\,=(2^2)(3^2)(4^2)\)
\(\,\,\,\,\,=(4)(9)(16)\)
\(\,\,\,\,\,=576\)
\(\,\,\,\,\,\displaystyle \prod_{i=2}^{4} i^2\)
\(\,\,\,\,\,=(2^2)(3^2)(4^2)\)
\(\,\,\,\,\,=(4)(9)(16)\)
\(\,\,\,\,\,=576\)
\(\textbf{12)}\) \( \displaystyle \prod_{i=1}^{3} (4i-1) \)
The answer is \(231\)
\(\,\,\,\,\,\displaystyle \prod_{i=1}^{3} (4i-1)\)
\(\,\,\,\,\,=(3)(7)(11)\)
\(\,\,\,\,\,=231\)
\(\,\,\,\,\,\displaystyle \prod_{i=1}^{3} (4i-1)\)
\(\,\,\,\,\,=(3)(7)(11)\)
\(\,\,\,\,\,=231\)
\(\textbf{13)}\) \( \displaystyle \prod_{i=1}^{4} \frac{i+1}{i} \)
The answer is \(5\)
\(\,\,\,\,\,\displaystyle \prod_{i=1}^{4} \frac{i+1}{i}\)
\(\,\,\,\,\,=\frac{2}{1}\cdot\frac{3}{2}\cdot\frac{4}{3}\cdot\frac{5}{4}\)
\(\,\,\,\,\,=5\)
\(\,\,\,\,\,\displaystyle \prod_{i=1}^{4} \frac{i+1}{i}\)
\(\,\,\,\,\,=\frac{2}{1}\cdot\frac{3}{2}\cdot\frac{4}{3}\cdot\frac{5}{4}\)
\(\,\,\,\,\,=5\)
\(\textbf{14)}\) \( \displaystyle \prod_{i=1}^{3} (-i) \)
The answer is \(-6\)
\(\,\,\,\,\,\displaystyle \prod_{i=1}^{3} (-i)\)
\(\,\,\,\,\,=(-1)(-2)(-3)\)
\(\,\,\,\,\,=-6\)
\(\,\,\,\,\,\displaystyle \prod_{i=1}^{3} (-i)\)
\(\,\,\,\,\,=(-1)(-2)(-3)\)
\(\,\,\,\,\,=-6\)
\(\textbf{15)}\) \( \displaystyle \prod_{i=2}^{5} (i-1) \)
The answer is \(24\)
\(\,\,\,\,\,\displaystyle \prod_{i=2}^{5} (i-1)\)
\(\,\,\,\,\,=(1)(2)(3)(4)\)
\(\,\,\,\,\,=24\)
\(\,\,\,\,\,\displaystyle \prod_{i=2}^{5} (i-1)\)
\(\,\,\,\,\,=(1)(2)(3)(4)\)
\(\,\,\,\,\,=24\)
\(\textbf{16)}\) \( \displaystyle \prod_{i=1}^{5} \frac{i}{i+1} \)
The answer is \(\frac{1}{6}\)
\(\,\,\,\,\,\displaystyle \prod_{i=1}^{5} \frac{i}{i+1}\)
\(\,\,\,\,\,=\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot\frac{4}{5}\cdot\frac{5}{6}\)
\(\,\,\,\,\,=\frac{1}{6}\)
\(\,\,\,\,\,\displaystyle \prod_{i=1}^{5} \frac{i}{i+1}\)
\(\,\,\,\,\,=\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot\frac{4}{5}\cdot\frac{5}{6}\)
\(\,\,\,\,\,=\frac{1}{6}\)
\(\textbf{17)}\) \( \displaystyle \prod_{i=1}^{4} (2^i-1) \)
The answer is \(315\)
\(\,\,\,\,\,\displaystyle \prod_{i=1}^{4} (2^i-1)\)
\(\,\,\,\,\,=(2^1-1)(2^2-1)(2^3-1)(2^4-1)\)
\(\,\,\,\,\,=(1)(3)(7)(15)\)
\(\,\,\,\,\,=315\)
\(\,\,\,\,\,\displaystyle \prod_{i=1}^{4} (2^i-1)\)
\(\,\,\,\,\,=(2^1-1)(2^2-1)(2^3-1)(2^4-1)\)
\(\,\,\,\,\,=(1)(3)(7)(15)\)
\(\,\,\,\,\,=315\)
\(\textbf{18)}\) \( \displaystyle \prod_{i=1}^{4} \frac{2i}{i+1} \)
The answer is \(\frac{16}{5}\)
\(\,\,\,\,\,\displaystyle \prod_{i=1}^{4} \frac{2i}{i+1}\)
\(\,\,\,\,\,=\frac{2}{2}\cdot\frac{4}{3}\cdot\frac{6}{4}\cdot\frac{8}{5}\)
\(\,\,\,\,\,=1\cdot\frac{4}{3}\cdot\frac{3}{2}\cdot\frac{8}{5}\)
\(\,\,\,\,\,=\frac{16}{5}\)
\(\,\,\,\,\,\displaystyle \prod_{i=1}^{4} \frac{2i}{i+1}\)
\(\,\,\,\,\,=\frac{2}{2}\cdot\frac{4}{3}\cdot\frac{6}{4}\cdot\frac{8}{5}\)
\(\,\,\,\,\,=1\cdot\frac{4}{3}\cdot\frac{3}{2}\cdot\frac{8}{5}\)
\(\,\,\,\,\,=\frac{16}{5}\)
\(\textbf{19)}\) 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 \(\frac{256}{175}\).
\(\,\,\,\,\,\displaystyle \prod_{n=1}^{3}\frac{(2n)(2n)}{(2n-1)(2n+1)}\)
\(\,\,\,\,\,=\frac{(2)(2)}{(1)(3)}\cdot\frac{(4)(4)}{(3)(5)}\cdot\frac{(6)(6)}{(5)(7)}\)
\(\,\,\,\,\,=\frac{4}{3}\cdot\frac{16}{15}\cdot\frac{36}{35}\)
\(\,\,\,\,\,=\frac{2304}{1575}\)
\(\,\,\,\,\,=\frac{256}{175}\)
\(\,\,\,\,\,\displaystyle \prod_{n=1}^{3}\frac{(2n)(2n)}{(2n-1)(2n+1)}\)
\(\,\,\,\,\,=\frac{(2)(2)}{(1)(3)}\cdot\frac{(4)(4)}{(3)(5)}\cdot\frac{(6)(6)}{(5)(7)}\)
\(\,\,\,\,\,=\frac{4}{3}\cdot\frac{16}{15}\cdot\frac{36}{35}\)
\(\,\,\,\,\,=\frac{2304}{1575}\)
\(\,\,\,\,\,=\frac{256}{175}\)
\(\textbf{20)}\)\(\displaystyle \prod_{n=2}^{\infty}\frac{n^3-1}{n^3+1}\)
The answer is \(\frac{2}{3}\).
\(\,\,\,\,\,\displaystyle \prod_{n=2}^{\infty}\frac{n^3-1}{n^3+1}\)
\(\,\,\,\,\,\displaystyle n^3-1=(n-1)(n^2+n+1)\)
\(\,\,\,\,\,\displaystyle n^3+1=(n+1)(n^2-n+1)\)
\(\,\,\,\,\,\displaystyle \frac{n^3-1}{n^3+1}=\frac{(n-1)(n^2+n+1)}{(n+1)(n^2-n+1)}\)
\(\,\,\,\,\,\displaystyle n^2+n+1=(n+1)^2-(n+1)+1\)
\(\,\,\,\,\,\displaystyle \prod_{n=2}^{N}\frac{n^3-1}{n^3+1}=\prod_{n=2}^{N}\frac{n-1}{n+1}\cdot\prod_{n=2}^{N}\frac{n^2+n+1}{n^2-n+1}\)
\(\,\,\,\,\,\displaystyle \prod_{n=2}^{N}\frac{n-1}{n+1}=\frac{1\cdot2\cdot3\cdots(N-1)}{3\cdot4\cdot5\cdots(N+1)}=\frac{2}{N(N+1)}\)
\(\,\,\,\,\,\displaystyle \prod_{n=2}^{N}\frac{n^2+n+1}{n^2-n+1}=\frac{N^2+N+1}{3}\)
\(\,\,\,\,\,\displaystyle \prod_{n=2}^{N}\frac{n^3-1}{n^3+1}=\frac{2}{N(N+1)}\cdot\frac{N^2+N+1}{3}\)
\(\,\,\,\,\,\displaystyle \prod_{n=2}^{N}\frac{n^3-1}{n^3+1}=\frac{2(N^2+N+1)}{3N(N+1)}\)
\(\,\,\,\,\,\displaystyle \prod_{n=2}^{\infty}\frac{n^3-1}{n^3+1}=\lim_{N\to\infty}\frac{2(N^2+N+1)}{3N(N+1)}\)
\(\,\,\,\,\,\displaystyle =\frac{2}{3}\)
\(\,\,\,\,\,\displaystyle \prod_{n=2}^{\infty}\frac{n^3-1}{n^3+1}\)
\(\,\,\,\,\,\displaystyle n^3-1=(n-1)(n^2+n+1)\)
\(\,\,\,\,\,\displaystyle n^3+1=(n+1)(n^2-n+1)\)
\(\,\,\,\,\,\displaystyle \frac{n^3-1}{n^3+1}=\frac{(n-1)(n^2+n+1)}{(n+1)(n^2-n+1)}\)
\(\,\,\,\,\,\displaystyle n^2+n+1=(n+1)^2-(n+1)+1\)
\(\,\,\,\,\,\displaystyle \prod_{n=2}^{N}\frac{n^3-1}{n^3+1}=\prod_{n=2}^{N}\frac{n-1}{n+1}\cdot\prod_{n=2}^{N}\frac{n^2+n+1}{n^2-n+1}\)
\(\,\,\,\,\,\displaystyle \prod_{n=2}^{N}\frac{n-1}{n+1}=\frac{1\cdot2\cdot3\cdots(N-1)}{3\cdot4\cdot5\cdots(N+1)}=\frac{2}{N(N+1)}\)
\(\,\,\,\,\,\displaystyle \prod_{n=2}^{N}\frac{n^2+n+1}{n^2-n+1}=\frac{N^2+N+1}{3}\)
\(\,\,\,\,\,\displaystyle \prod_{n=2}^{N}\frac{n^3-1}{n^3+1}=\frac{2}{N(N+1)}\cdot\frac{N^2+N+1}{3}\)
\(\,\,\,\,\,\displaystyle \prod_{n=2}^{N}\frac{n^3-1}{n^3+1}=\frac{2(N^2+N+1)}{3N(N+1)}\)
\(\,\,\,\,\,\displaystyle \prod_{n=2}^{\infty}\frac{n^3-1}{n^3+1}=\lim_{N\to\infty}\frac{2(N^2+N+1)}{3N(N+1)}\)
\(\,\,\,\,\,\displaystyle =\frac{2}{3}\)
See Related Pages
\(\bullet\text{ Product Notation Calculator }\)
\(\,\,\,\,\,\,\,\,\text{(Symbolab.com)}\)
\(\bullet\text{ Summation Notation (Sigma Notation)}\)
\(\,\,\,\,\,\,\,\displaystyle \sum_{i=3}^{5} 3-2i\)