Math B Sigma Notation and Sums: Practice Problems Practise the sigma formulas and the standard tricks for sums: splitting into differences, grouping terms, and computing S − r S S - rS S − r S
Basic, Standard: Grade 11 · Term 2 / Advanced: Grade 12 · Exam prep
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Guide Examples Practice Related
Difference sequences The sequence b n = a n + 1 − a n b_n = a_{n+1} - a_n b n = a n + 1 − a n difference sequence of { a n } \{a_n\} { a n }
Check
This holds for n ≧ 2 n \geqq 2 n ≧ 2 a 1 a_1 a 1 n = 1 n = 1 n = 1
Telescoping sums Write each term as a difference of two fractions ; then neighbouring terms cancel.
∑ k = 1 n 1 k ( k + 1 ) = ∑ k = 1 n ( 1 k − 1 k + 1 ) = 1 − 1 n + 1 = n n + 1 \sum_{k=1}^{n} \frac{1}{k(k+1)} = \sum_{k=1}^{n}\left(\frac{1}{k} - \frac{1}{k+1}\right) = 1 - \frac{1}{n+1} = \frac{n}{n+1} k = 1 ∑ n k ( k + 1 ) 1 = k = 1 ∑ n ( k 1 − k + 1 1 ) = 1 − n + 1 1 = n + 1 n Remember the factor in front: 1 ( 2 k − 1 ) ( 2 k + 1 ) = 1 2 ( 1 2 k − 1 − 1 2 k + 1 ) \frac{1}{(2k-1)(2k+1)} = \frac{1}{2}\left(\frac{1}{2k-1} - \frac{1}{2k+1}\right) ( 2 k − 1 ) ( 2 k + 1 ) 1 = 2 1 ( 2 k − 1 1 − 2 k + 1 1 )
Grouped sequences and S − rS In a grouped sequence where group n n n n n n 1 1 1 n − 1 n-1 n − 1 1 2 n ( n − 1 ) \frac{1}{2}n(n-1) 2 1 n ( n − 1 ) n n n
For sums like S = 1 + 2 ⋅ 2 + 3 ⋅ 2 2 + ⋯ + n ⋅ 2 n − 1 S = 1 + 2 \cdot 2 + 3 \cdot 2^2 + \cdots + n \cdot 2^{n-1} S = 1 + 2 ⋅ 2 + 3 ⋅ 2 2 + ⋯ + n ⋅ 2 n − 1 S − 2 S S - 2S S − 2 S
S − 2 S = 1 + 2 + 2 2 + ⋯ + 2 n − 1 − n ⋅ 2 n ⟹ S = ( n − 1 ) ⋅ 2 n + 1 S - 2S = 1 + 2 + 2^2 + \cdots + 2^{n-1} - n \cdot 2^n \quad\Longrightarrow\quad S = (n-1) \cdot 2^n + 1 S − 2 S = 1 + 2 + 2 2 + ⋯ + 2 n − 1 − n ⋅ 2 n ⟹ S = ( n − 1 ) ⋅ 2 n + 1 Worked examples
Find the sum.
∑ k = 1 10 k 3 \sum_{k=1}^{10} k^3 k = 1 ∑ 10 k 3 Check answer Hint Answer Solution + Worksheet ✓ Added
Hint Use the formulas with the given value of n n n
Solution Express the sum from k = 1 k = 1 k = 1 n n n n n n
∑ k = 1 n k 3 = 1 4 n 2 ( n + 1 ) 2 \sum_{k=1}^{n} k^3 = \frac{1}{4}n^{2}(n + 1)^{2} k = 1 ∑ n k 3 = 4 1 n 2 ( n + 1 ) 2 Substitute n = 10 n = 10 n = 10
Find the sum.
∑ k = 1 n 1 ( 2 k − 1 ) ( 2 k + 1 ) \sum_{k=1}^{n} \frac{1}{(2k - 1)(2k + 1)} k = 1 ∑ n ( 2 k − 1 ) ( 2 k + 1 ) 1 Check answer Hint Answer Solution + Worksheet ✓ Added
Hint Use partial fractions; neighbouring terms cancel.
Answer n 2 n + 1 \frac{n}{2n + 1} 2 n + 1 n Solution Use partial fractions.
1 ( 2 k − 1 ) ( 2 k + 1 ) = 1 2 ( 1 2 k − 1 − 1 2 k + 1 ) \frac{1}{(2k - 1)(2k + 1)} = \frac{1}{2}\left(\frac{1}{2k - 1} - \frac{1}{2k + 1}\right) ( 2 k − 1 ) ( 2 k + 1 ) 1 = 2 1 ( 2 k − 1 1 − 2 k + 1 1 ) Summing, the middle terms cancel.
1 2 { ( 1 − 1 3 ) + ( 1 3 − 1 5 ) + ⋯ + ( 1 2 n − 1 − 1 2 n + 1 ) } \frac{1}{2}\left\{\left(1 - \frac{1}{3}\right) + \left(\frac{1}{3} - \frac{1}{5}\right) + \cdots + \left(\frac{1}{2n - 1} - \frac{1}{2n + 1}\right)\right\} 2 1 { ( 1 − 3 1 ) + ( 3 1 − 5 1 ) + ⋯ + ( 2 n − 1 1 − 2 n + 1 1 ) } Hence
1 2 ( 1 − 1 2 n + 1 ) = n 2 n + 1 \begin{aligned}\frac{1}{2}\left(1 - \frac{1}{2n + 1}\right) &= \frac{n}{2n + 1}\end{aligned} 2 1 ( 1 − 2 n + 1 1 ) = 2 n + 1 n
Find the sum.
∑ k = 1 n 1 k + 1 + k \sum_{k=1}^{n} \frac{1}{\sqrt{k + 1} + \sqrt{k}} k = 1 ∑ n k + 1 + k 1 Check answer Hint Answer Solution + Worksheet ✓ Added
Hint Rationalize the denominator to get a difference.
Answer n + 1 − 1 \sqrt{n + 1} - 1 n + 1 − 1 Solution Rationalize the denominator.
1 k + 1 + k = k + 1 − k \frac{1}{\sqrt{k + 1} + \sqrt{k}} = \sqrt{k + 1} - \sqrt{k} k + 1 + k 1 = k + 1 − k Summing, the middle terms cancel.
( 2 − 1 ) + ( 3 − 2 ) + ⋯ + ( n + 1 − n ) = n + 1 − 1 (\sqrt{2} - \sqrt{1}) + (\sqrt{3} - \sqrt{2}) + \cdots + (\sqrt{n + 1} - \sqrt{n}) = \sqrt{n + 1} - 1 ( 2 − 1 ) + ( 3 − 2 ) + ⋯ + ( n + 1 − n ) = n + 1 − 1 Practice problems
Find the sum.
∑ k = 1 11 k 3 \sum_{k=1}^{11} k^3 k = 1 ∑ 11 k 3 Check answer Hint Answer Solution + Worksheet ✓ Added
Hint Use the formulas with the given value of n n n
Solution Express the sum from k = 1 k = 1 k = 1 n n n n n n
∑ k = 1 n k 3 = 1 4 n 2 ( n + 1 ) 2 \sum_{k=1}^{n} k^3 = \frac{1}{4}n^{2}(n + 1)^{2} k = 1 ∑ n k 3 = 4 1 n 2 ( n + 1 ) 2 Substitute n = 11 n = 11 n = 11
Find the sum.
∑ k = 1 n 1 ( 3 k + 2 ) ( 3 k + 5 ) \sum_{k=1}^{n} \frac{1}{(3k + 2)(3k + 5)} k = 1 ∑ n ( 3 k + 2 ) ( 3 k + 5 ) 1 Check answer Hint Answer Solution + Worksheet ✓ Added
Hint Use partial fractions; neighbouring terms cancel.
Answer n 5 ( 3 n + 5 ) \frac{n}{5(3n + 5)} 5 ( 3 n + 5 ) n Solution Use partial fractions.
1 ( 3 k + 2 ) ( 3 k + 5 ) = 1 3 ( 1 3 k + 2 − 1 3 k + 5 ) \frac{1}{(3k + 2)(3k + 5)} = \frac{1}{3}\left(\frac{1}{3k + 2} - \frac{1}{3k + 5}\right) ( 3 k + 2 ) ( 3 k + 5 ) 1 = 3 1 ( 3 k + 2 1 − 3 k + 5 1 ) Summing, the middle terms cancel.
1 3 { ( 1 5 − 1 8 ) + ( 1 8 − 1 11 ) + ⋯ + ( 1 3 n + 2 − 1 3 n + 5 ) } \frac{1}{3}\left\{\left(\frac{1}{5} - \frac{1}{8}\right) + \left(\frac{1}{8} - \frac{1}{11}\right) + \cdots + \left(\frac{1}{3n + 2} - \frac{1}{3n + 5}\right)\right\} 3 1 { ( 5 1 − 8 1 ) + ( 8 1 − 11 1 ) + ⋯ + ( 3 n + 2 1 − 3 n + 5 1 ) } Hence
1 3 ( 1 5 − 1 3 n + 5 ) = n 5 ( 3 n + 5 ) \begin{aligned}\frac{1}{3}\left(\frac{1}{5} - \frac{1}{3n + 5}\right) &= \frac{n}{5(3n + 5)}\end{aligned} 3 1 ( 5 1 − 3 n + 5 1 ) = 5 ( 3 n + 5 ) n
Find the sum.
∑ k = 1 n 1 k ( k + 2 ) \sum_{k=1}^{n} \frac{1}{k(k + 2)} k = 1 ∑ n k ( k + 2 ) 1 Check answer Hint Answer Solution + Worksheet ✓ Added
Hint With partial fractions, terms two apart cancel.
Answer n ( 3 n + 5 ) 4 ( n + 1 ) ( n + 2 ) \frac{n(3n + 5)}{4(n + 1)(n + 2)} 4 ( n + 1 ) ( n + 2 ) n ( 3 n + 5 ) Solution Use partial fractions.
1 k ( k + 2 ) = 1 2 ( 1 k − 1 k + 2 ) \frac{1}{k(k + 2)} = \frac{1}{2}\left(\frac{1}{k} - \frac{1}{k + 2}\right) k ( k + 2 ) 1 = 2 1 ( k 1 − k + 2 1 ) Summing, only these terms survive.
1 2 ( 1 + 1 2 − 1 n + 1 − 1 n + 2 ) = n ( 3 n + 5 ) 4 ( n + 1 ) ( n + 2 ) \begin{aligned}\frac{1}{2}\left(1 + \frac{1}{2} - \frac{1}{n + 1} - \frac{1}{n + 2}\right) &= \frac{n(3n + 5)}{4(n + 1)(n + 2)}\end{aligned} 2 1 ( 1 + 2 1 − n + 1 1 − n + 2 1 ) = 4 ( n + 1 ) ( n + 2 ) n ( 3 n + 5 )