14 exam-style questions on standard results, method of differences and proofs. Tap Show answer for full working.
Use the standard results for \(\sum r\) and \(\sum r^2\) to find \(\displaystyle\sum_{r=1}^{n} (3r^2 - 2r)\).
Simplify your answer fully.
\(\sum_{r=1}^{n}(3r^2-2r) = 3\cdot\dfrac{n(n+1)(2n+1)}{6} - 2\cdot\dfrac{n(n+1)}{2}\)
\(= \dfrac{n(n+1)(2n+1)}{2} - n(n+1) = n(n+1)\left(\dfrac{2n+1}{2} - 1\right)\)
\(= n(n+1)\left(\dfrac{2n+1-2}{2}\right) = \dfrac{n(n+1)(2n-1)}{2}\)
Show that \(\dfrac{1}{r(r+1)} = \dfrac{1}{r} - \dfrac{1}{r+1}\).
Hence find \(\displaystyle\sum_{r=1}^{n} \dfrac{1}{r(r+1)}\).
RHS: \(\dfrac{1}{r} - \dfrac{1}{r+1} = \dfrac{r+1-r}{r(r+1)} = \dfrac{1}{r(r+1)}\). ✓
Sum: \(\sum_{r=1}^{n}\left(\dfrac{1}{r}-\dfrac{1}{r+1}\right) = \left(1-\dfrac{1}{2}\right)+\left(\dfrac{1}{2}-\dfrac{1}{3}\right)+\cdots+\left(\dfrac{1}{n}-\dfrac{1}{n+1}\right)\)
\(= 1 - \dfrac{1}{n+1} = \dfrac{n}{n+1}\)
Find \(\displaystyle\sum_{r=1}^{20} r(r+1)\).
\(\sum r(r+1) = \sum(r^2+r) = \sum r^2 + \sum r\)
\(= \dfrac{n(n+1)(2n+1)}{6} + \dfrac{n(n+1)}{2}\) with \(n=20\)
\(= \dfrac{20\cdot21\cdot41}{6} + \dfrac{20\cdot21}{2} = 2870 + 210 = 3080\)
Alternatively: \(\sum r(r+1) = \sum(r^2+r) = \dfrac{n(n+1)(n+2)}{3} = \dfrac{20\cdot21\cdot22}{3} = 3080\).
Given that \(\dfrac{2r+1}{r(r+1)(r+2)} = \dfrac{a}{r} + \dfrac{b}{r+1} + \dfrac{c}{r+2}\), find the values of \(a\), \(b\) and \(c\).
Hence find \(\displaystyle\sum_{r=1}^{n} \dfrac{2r+1}{r(r+1)(r+2)}\).
\(2r+1 = a(r+1)(r+2) + b r(r+2) + c r(r+1)\)
Put \(r=0\): \(1 = a\cdot1\cdot2 \Rightarrow a = \dfrac{1}{2}\)
Put \(r=-1\): \(-1 = b(-1)(1) \Rightarrow -1 = -b \Rightarrow b = 1\)
Put \(r=-2\): \(-3 = c(-2)(-1) \Rightarrow -3 = 2c \Rightarrow c = -\dfrac{3}{2}\)
Check coeff of \(r^2\): \(0 = a+b+c = \frac12+1-\frac32=0\). ✓
Sum: \(\sum\left(\dfrac{1/2}{r} + \dfrac{1}{r+1} - \dfrac{3/2}{r+2}\right)\). Method of differences gives
\(\dfrac{n(3n+1)}{2(n+1)(n+2)}\) (after simplifying the telescoping sum).
Use the standard formulae to find \(\displaystyle\sum_{r=1}^{n} r(3r-1)\).
Hence find \(\displaystyle\sum_{r=10}^{20} r(3r-1)\).
\(\sum r(3r-1) = 3\sum r^2 - \sum r = 3\cdot\dfrac{n(n+1)(2n+1)}{6} - \dfrac{n(n+1)}{2}\)
\(= \dfrac{n(n+1)(2n+1)}{2} - \dfrac{n(n+1)}{2} = \dfrac{n(n+1)}{2}(2n+1-1) = n^2(n+1)\)
\(\sum_{r=10}^{20} = \sum_{1}^{20} - \sum_{1}^{9} = 20^2\cdot21 - 9^2\cdot10 = 8400 - 810 = 7590\)
Prove by induction that \(\displaystyle\sum_{r=1}^{n} r\cdot 2^{r-1} = (n-1)2^n + 1\) for all positive integers \(n\).
Base: \(n=1\): LHS \(= 1\cdot 2^0 = 1\). RHS \(= (0)2^1 + 1 = 1\). ✓
Assume true for \(n=k\): \(\sum_{r=1}^{k} r\cdot 2^{r-1} = (k-1)2^k + 1\).
Inductive step: \(\sum_{r=1}^{k+1} r\cdot 2^{r-1} = (k-1)2^k + 1 + (k+1)2^k\)
\(= (k-1+k+1)2^k + 1 = (2k)2^k + 1 = k\cdot 2^{k+1} + 1\)
\(= ((k+1)-1)2^{k+1} + 1\). ✓
By induction true for all positive integers \(n\).
Find \(\displaystyle\sum_{r=1}^{n} \dfrac{1}{(2r-1)(2r+1)}\) using partial fractions.
\(\dfrac{1}{(2r-1)(2r+1)} = \dfrac{1/2}{2r-1} - \dfrac{1/2}{2r+1} = \dfrac{1}{2}\left(\dfrac{1}{2r-1}-\dfrac{1}{2r+1}\right)\)
Sum telescopes: \(\dfrac{1}{2}\Big[\left(1-\dfrac{1}{3}\right)+\left(\dfrac{1}{3}-\dfrac{1}{5}\right)+\cdots+\left(\dfrac{1}{2n-1}-\dfrac{1}{2n+1}\right)\Big]\)
\(= \dfrac{1}{2}\left(1 - \dfrac{1}{2n+1}\right) = \dfrac{n}{2n+1}\)
Show that \(\dfrac{r}{(r+1)(r+2)(r+3)} = \dfrac{a}{r+1} + \dfrac{b}{r+2} + \dfrac{c}{r+3}\) and find \(a\), \(b\), \(c\).
Hence find \(\displaystyle\sum_{r=1}^{n} \dfrac{r}{(r+1)(r+2)(r+3)}\) and deduce the sum to infinity.
\(r = a(r+2)(r+3) + b(r+1)(r+3) + c(r+1)(r+2)\)
\(r=-1\): \(-1 = a(1)(2) \Rightarrow a = -\dfrac{1}{2}\)
\(r=-2\): \(-2 = b(-1)(1) \Rightarrow -2 = -b \Rightarrow b = 2\)
\(r=-3\): \(-3 = c(-2)(-1) \Rightarrow -3 = 2c \Rightarrow c = -\dfrac{3}{2}\)
Sum: \(\sum\left(-\dfrac{1/2}{r+1} + \dfrac{2}{r+2} - \dfrac{3/2}{r+3}\right)\). Telescoping yields
\(\dfrac{n(n+1)}{4(n+2)(n+3)}\). As \(n\to\infty\), sum \(\to \dfrac{1}{4}\).
Using the formula for \(\sum r^3\), show that \(\displaystyle\sum_{r=1}^{n} r^2(r+1) = \dfrac{n(n+1)(n+2)(3n+1)}{12}\).
\(\sum r^2(r+1) = \sum(r^3+r^2) = \sum r^3 + \sum r^2\)
\(= \left(\dfrac{n(n+1)}{2}\right)^2 + \dfrac{n(n+1)(2n+1)}{6}\)
\(= \dfrac{n^2(n+1)^2}{4} + \dfrac{n(n+1)(2n+1)}{6}\)
\(= \dfrac{n(n+1)}{12}\big[3n(n+1) + 2(2n+1)\big] = \dfrac{n(n+1)}{12}(3n^2+3n+4n+2)\)
\(= \dfrac{n(n+1)}{12}(3n^2+7n+2) = \dfrac{n(n+1)(3n+1)(n+2)}{12}\). ✓
Prove by induction that \(\displaystyle\sum_{r=1}^{n} \dfrac{1}{r(r+1)} = \dfrac{n}{n+1}\) for all positive integers \(n\).
Hence find \(\displaystyle\sum_{r=n+1}^{2n} \dfrac{1}{r(r+1)}\).
Base \(n=1\): LHS \(= \dfrac{1}{1\cdot2} = \dfrac{1}{2}\). RHS \(= \dfrac{1}{2}\). ✓
Assume for \(n=k\): \(\sum_{r=1}^{k}\dfrac{1}{r(r+1)} = \dfrac{k}{k+1}\).
Step: \(\sum_{1}^{k+1} = \dfrac{k}{k+1} + \dfrac{1}{(k+1)(k+2)} = \dfrac{k(k+2)+1}{(k+1)(k+2)} = \dfrac{k^2+2k+1}{(k+1)(k+2)} = \dfrac{(k+1)^2}{(k+1)(k+2)} = \dfrac{k+1}{k+2}\). ✓
\(\sum_{r=n+1}^{2n} = \sum_{1}^{2n} - \sum_{1}^{n} = \dfrac{2n}{2n+1} - \dfrac{n}{n+1} = \dfrac{2n(n+1) - n(2n+1)}{(2n+1)(n+1)} = \dfrac{n}{(2n+1)(n+1)}\).
Find \( \displaystyle\sum_{r=1}^{n} (2r + 1) \) using the standard formulae, and simplify.
\( \sum (2r + 1) = 2\sum r + \sum 1 = 2\cdot\dfrac{n(n+1)}{2} + n = n(n+1) + n = n^2 + 2n \).
Use the method of differences to show that \( \displaystyle\sum_{r=1}^{n} \dfrac{1}{r(r+1)} = 1 - \dfrac{1}{n+1} \).
Hence find \( \displaystyle\sum_{r=1}^{\infty} \dfrac{1}{r(r+1)} \).
\( \dfrac{1}{r(r+1)} = \dfrac{1}{r} - \dfrac{1}{r+1} \) (partial fractions).
Sum telescopes: \( \left(1 - \dfrac{1}{2}\right) + \left(\dfrac{1}{2} - \dfrac{1}{3}\right) + \cdots + \left(\dfrac{1}{n} - \dfrac{1}{n+1}\right) = 1 - \dfrac{1}{n+1} \).
As \( n\to\infty \), the infinite sum is 1.
Find a closed form for \( \displaystyle\sum_{r=1}^{n} r\cdot 2^r \).
(You may use the formula for the sum of a geometric series.)
Let \( S = \sum_{r=1}^{n} r x^r \) with \( x=2 \). Then \( S = x + 2x^2 + \cdots + n x^n \).
\( xS = x^2 + 2x^3 + \cdots + (n-1)x^n + n x^{n+1} \).
\( S - xS = x + x^2 + \cdots + x^n - n x^{n+1} = x\dfrac{1-x^n}{1-x} - n x^{n+1} \).
\( S(1-x) = x\dfrac{1-x^n}{1-x} - n x^{n+1} \), so \( S = \dfrac{x(1-x^n)}{(1-x)^2} - \dfrac{n x^{n+1}}{1-x} \).
For \( x=2 \): \( S = \dfrac{2(1-2^n)}{(1-2)^2} - \dfrac{n\cdot 2^{n+1}}{1-2} = 2(1-2^n) + n\cdot 2^{n+1} = 2 - 2^{n+1} + n\cdot 2^{n+1} = 2 + (n-1)2^{n+1} \).
Prove by induction that \( \displaystyle\sum_{r=1}^{n} r^3 = \left( \dfrac{n(n+1)}{2} \right)^2 \) for all positive integers \( n \).
Base case \( n=1 \): LHS \( =1 \), RHS \( =(1\cdot2/2)^2=1 \). True.
Inductive hypothesis: assume true for \( n=k \): \( \sum_{r=1}^{k} r^3 = \left( \dfrac{k(k+1)}{2} \right)^2 \).
Inductive step: for \( n=k+1 \),
\( \sum_{r=1}^{k+1} r^3 = \left( \dfrac{k(k+1)}{2} \right)^2 + (k+1)^3 = (k+1)^2 \left[ \dfrac{k^2}{4} + (k+1) \right] = (k+1)^2 \left( \dfrac{k^2 + 4k + 4}{4} \right) = (k+1)^2 \cdot \dfrac{(k+2)^2}{4} = \left( \dfrac{(k+1)(k+2)}{2} \right)^2 \).
Hence true for all \( n \) by induction.