Sums

Exercises · Sums and products · with worked solutions · Lecture notes (PDF) · Chapter 1: Sums

Exercise 1

Prove, using the properties of the sums, that for every natural number \(n \ge 1\) we have:

\[\begin{equation} \label{EXSQ} \sum_{k=1}^{n} k^2 = \underbrace{\frac{2\;n^3 + 3\;n^2 + n}{6}}_{ =\frac{n\:(n+1)\:(2\:n+1)}{6}} \end{equation}\]

(sum of the squares of the first \(n\) natural numbers without zero).

Solution

We start by writing \(\sum_{k=0}^n (k+1)^3\) in two different ways:

\[\begin{align*} 1)~~~\sum_{k=0}^n (k+1)^3 &=\sum_{k=0}^n(k^3+3\:k^2+3\:k+1)\\[2ex] &= \left( \sum_{k=1}^n k^3 + 3\: \sum_{k=1}^n k^2 + 3\: \sum_{k=1}^n k + \sum_{k=1}^n 1 \right) +1 \\[2ex] &=\sum_{k=1}^n k^3 + 3\: \sum_{k=1}^n k^2 + \frac{3\;(n^2+n)}{2} +n +1\\[5ex] 2)~~~ \sum_{k=0}^n (k+1)^3&=\sum_{k=1}^{n+1} k^3=\sum_{k=1}^n k^3 + (n+1)^3 \end{align*}\]

(in 1) the term with \(k=0\), equal to \(1\), has been separated from the others). Equating the expressions and canceling the sum with \(k^3\) we get:

\[ 3\: \sum_{k=1}^n k^2 + \frac{3\:(n^2+n)}{2} +n +1= (n+1)^3 \]

Isolating the sum with \(k^2\) we get:

\[\begin{align*} 3\: \sum_{k=1}^n k^2 &= (n+1)^3 - \frac{3\:(n^2+n)}{2} -n -1 \\[2ex] &= n^3 + 3n^2 +3n+1 - \frac{3n^2+3n}{2} -n -1\\[2ex] &= \frac{2n^3 + 3n^2 + n}{2} \end{align*}\]

Then we have:

\[\begin{align*} \sum_{k=1}^n k^2 &= \frac{2n^3 + 3n^2 + n}{6} \end{align*}\]

Exercise 2

Compute the sum of the squares of the first \(10\), \(100\) and \(1000\) natural numbers (without zero).

Solution

Using formula \(\eqref{EXSQ}\), the sum of squares of the first 10 natural numbers (without zero) is:

\[ \sum_{k=1}^{10} k^2 = \frac{2\cdot 10^3 + 3\cdot 10^2 + 10}{6} = 385 \]

The sum of squares of the first 100 natural numbers (without zero) is:

\[ \sum_{k=1}^{100} k^2 = \frac{2\cdot 100^3 + 3\cdot 100^2 + 100}{6} = 338.350 \]

The sum of squares of the first 1000 natural numbers (without zero) is:

\[ \sum_{k=1}^{1000} k^2 = \frac{2\cdot 1000^3 + 3\cdot 1000^2 + 1000}{6} = 333.833.500 \]

Exercise 3

Prove, using the properties of the sums, that for every natural number \(n \ge 1\) we have:

\[\begin{equation} \label{EXCU} \sum_{k=1}^{n} k^3 = \underbrace{\frac{n^4 + 2\;n^3 + n^2}{4}}_{ =\frac{n^2\:(n+1)^2}{4}=\frac{(n^2+n)^2}{4}} \end{equation}\]

(sum of the cubes of the first \(n\) natural numbers without zero).

Solution

We start by writing \(\sum_{k=0}^n (k+1)^4\) in two different ways, using formula \(\eqref{EXSQ}\) for the sum of the squares:

\[\begin{align*} 1)~~~\sum_{k=0}^n (k+1)^4 &=\sum_{k=0}^n(k^4+4\:k^3+6\:k^2+4\:k+1)\\[2ex] &= \left( \sum_{k=1}^n k^4 + 4\: \sum_{k=1}^n k^3 + 6\: \sum_{k=1}^n k^2 + 4\: \sum_{k=1}^n k + \sum_{k=1}^n 1 \right) +1 \\[2ex] &=\sum_{k=1}^n k^4 + 4\: \sum_{k=1}^n k^3 + 2n^3 + 3n^2 + n + 2\:n^2 +2\:n +n +1\\[2ex] &=\sum_{k=1}^n k^4 + 4\: \sum_{k=1}^n k^3 + 2n^3 + 5n^2 +4n +1\\[2ex] 2)~~~ \sum_{k=0}^n (k+1)^4&=\sum_{k=1}^{n+1} k^4=\sum_{k=1}^n k^4 + (n+1)^4 \end{align*}\]

Equating the expressions and canceling the sum with \(k^4\) we get:

\[ 4\: \sum_{k=1}^n k^3 + 2n^3 + 5 n^2 +4n +1 = (n+1)^4 \]

Isolating the sum with \(k^3\) we get:

\[\begin{align*} 4\: \sum_{k=1}^n k^3 &= n^4 + 4n^3 +6n^2 + 4n +1 - 2n^3 - 5 n^2 -4n -1 = n^4 + 2n^3 + n^2 \end{align*}\]

Then we have:

\[\begin{align*} \sum_{k=1}^n k^3 & = \frac{n^4 + 2n^3 + n^2}{4} \end{align*}\]

Exercise 4

Compute the sum of the cubes of the first \(10\), \(100\) and \(1000\) natural numbers (without zero).

Solution

Using formula \(\eqref{EXCU}\), the sum of cubes of the first 10 natural numbers (without zero) is:

\[ \sum_{k=1}^{10} k^3 = \frac{ 10^4 + 2 \cdot 10^3 + 10^2}{4} = 3025 \]

The sum of cubes of the first 100 natural numbers (without zero) is:

\[ \sum_{k=1}^{100} k^3 = \frac{ 100^4 + 2 \cdot 100^3 + 100^2}{4} = 25.502.500 \]

The sum of cubes of the first 1000 natural numbers (without zero) is:

\[ \sum_{k=1}^{1000} k^3 = \frac{ 1000^4 + 2 \cdot 1000^3 + 1000^2}{4} = 250.500.250.000 \]

Exercise 5

Prove that the sum of the cubes of the first \(n\) natural numbers (without zero) equals the square of the sum of the first \(n\) natural numbers (without zero):

\[ \sum_{k=1}^{n} k^3 = \left(\sum_{k=1}^{n} k \right)^2 \]

Solution

Using formula \(\eqref{EXCU}\) and the formula \(\sum_{k=1}^{n} k = \frac{n\:(n+1)}{2}\), we have:

\[ \sum_{k=1}^{n} k^3 = \frac{n^4 + 2n^3 + n^2}{4} = \frac{n^2 \; (n^2 + 2n + 1)}{4} = \frac{n^2\:(n+1)^2}{4}= \left(\frac{n\:(n+1)}{2}\right)^2= \left(\sum_{k=1}^{n} k \right)^2 \]

Exercise 6

Using the index translation, rewrite the following sums so that the sum index starts from \(1\), and compute them:

\[ {\rm a)}~ \sum_{j=3}^{12} (j-2)^2, \qquad {\rm b)}~ \sum_{j=0}^{n-1} (j+1), \qquad {\rm c)}~ \sum_{j=0}^{49} (2\:j+1) \]

Solution

• a) With the index translation by \(m=2\) (setting \(i=j-2\), when \(j\) goes from \(3\) to \(12\), \(i\) goes from \(1\) to \(10\)) and formula \(\eqref{EXSQ}\):

  \[ \sum_{j=3}^{12} (j-2)^2 = \sum_{i=1}^{10} i^2 = \frac{2\cdot 10^3 + 3 \cdot 10^2 + 10}{6} = 385 \]

• b) With the index translation by \(m=1\) (setting \(i=j+1\)):

  \[ \sum_{j=0}^{n-1} (j+1) = \sum_{i=1}^{n} i = \frac{n^2+n}{2} \]

• c) With the index translation by \(m=1\) (setting \(i=j+1\), so that \(2\:j+1 = 2\:i-1\)), this is the sum of the first \(50\) odd numbers:

  \[ \sum_{j=0}^{49} (2\:j+1) = \sum_{i=1}^{50} (2\:i-1) = 50^2 = 2500 \]

Exercise 7

Find a closed formula for the sum

\[ \sum_{j=1}^{n} \big(3\:j^2 - 2\:j + 5\big) \]

and use it to compute the value for \(n=10\).

Solution

Using the union of sums, the product by a constant and the sum with constant term, we have:

\[\begin{align*} \sum_{j=1}^{n} \big(3\:j^2 - 2\:j + 5\big) &= 3\: \sum_{j=1}^{n} j^2 - 2\: \sum_{j=1}^{n} j + \sum_{j=1}^{n} 5 \\[2ex] &= 3 \: \frac{2n^3 + 3n^2 + n}{6} - 2\: \frac{n^2+n}{2} + 5\:n \\[2ex] &= n^3 + \frac{3}{2}\: n^2 + \frac{1}{2}\: n - n^2 - n + 5\:n = \frac{2n^3 + n^2 + 9n}{2} \end{align*}\]

For \(n=10\) we get:

\[ \sum_{j=1}^{10} \big(3\:j^2 - 2\:j + 5\big) = \frac{2000 + 100 + 90}{2} = 1095 \]

As a check: \(3 \cdot 385 - 2 \cdot 55 + 5 \cdot 10 = 1155 - 110 + 50 = 1095\).

Exercise 8

Using the decomposition property, compute:

\[ {\rm a)}~ \sum_{j=11}^{20} j^2, \qquad {\rm b)}~ \sum_{j=n+1}^{2n} j \quad (n \ge 1) \]

Solution

• a) From the decomposition \(\sum_{j=1}^{20} j^2 = \sum_{j=1}^{10} j^2 + \sum_{j=11}^{20} j^2\) and formula \(\eqref{EXSQ}\):

  \[ \sum_{j=11}^{20} j^2 = \sum_{j=1}^{20} j^2 - \sum_{j=1}^{10} j^2 = \frac{2\cdot 20^3 + 3 \cdot 20^2 + 20}{6} - 385 = 2870 - 385 = 2485 \]

• b) From the decomposition \(\sum_{j=1}^{2n} j = \sum_{j=1}^{n} j + \sum_{j=n+1}^{2n} j\):

  \[ \sum_{j=n+1}^{2n} j = \frac{(2n)^2 + 2n}{2} - \frac{n^2+n}{2} = \frac{4n^2 + 2n - n^2 - n}{2} = \frac{3n^2+n}{2} \]

Exercise 9

• a) Prove that for every sequence of real numbers \(a_1, a_2, \dots, a_{n+1}\) we have (telescoping sum):

  \[ \sum_{j=1}^{n} \big( a_{j+1} - a_j \big) = a_{n+1} - a_1 \]

• b) Using a), compute \(\displaystyle \sum_{j=1}^{n} \frac{1}{j\:(j+1)}\) and its value for \(n=99\).

Solution

• a) Using the union of sums, the index translation and the decomposition, we have:

  \[\begin{align*} \sum_{j=1}^{n} \big( a_{j+1} - a_j \big) &= \sum_{j=1}^{n} a_{j+1} - \sum_{j=1}^{n} a_j = \sum_{j=2}^{n+1} a_{j} - \sum_{j=1}^{n} a_j\\[2ex] &= \left(\sum_{j=2}^{n} a_{j} + a_{n+1}\right) - \left( a_1 + \sum_{j=2}^{n} a_j \right) = a_{n+1} - a_1 \end{align*}\]

• b) Since

  \[ \frac{1}{j\:(j+1)} = \frac{(j+1) - j}{j\:(j+1)} = \frac{1}{j} - \frac{1}{j+1} \]

  we can apply a) with \(a_j = -\frac{1}{j}\):

  \[ \sum_{j=1}^{n} \frac{1}{j\:(j+1)} = \sum_{j=1}^{n} \left( -\frac{1}{j+1} - \Big(-\frac{1}{j}\Big)\right) = -\frac{1}{n+1} + 1 = \frac{n}{n+1} \]

  For \(n=99\) we get \(\frac{99}{100}\).

Exercise 10

Compute \(\displaystyle \sum_{j=1}^{n} \big( (j+1)^2 - j^2 \big)\) in two different ways and deduce again the formula

\[ \sum_{j=1}^n j = \frac{n^2+n}{2} \]

Solution

First way: it is a telescoping sum (Exercise Exercise 9) with \(a_j = j^2\), then:

\[ \sum_{j=1}^{n} \big( (j+1)^2 - j^2 \big) = (n+1)^2 - 1 = n^2 + 2\:n \]

Second way: since \((j+1)^2 - j^2 = 2\:j + 1\), we have:

\[ \sum_{j=1}^{n} \big( (j+1)^2 - j^2 \big) = \sum_{j=1}^{n} (2\:j+1) = 2 \: \sum_{j=1}^{n} j + n \]

Equating the two expressions:

\[ 2 \: \sum_{j=1}^{n} j + n = n^2 + 2\:n {\rm ~~~~that~is~~~~} \sum_{j=1}^{n} j = \frac{n^2 + n}{2} \]

Exercise 11

A double sum \(\sum_{i=1}^{n} \sum_{j=1}^{m} a_{ij}\) is the sum for \(i\) from \(1\) to \(n\) of the inner sums \(\sum_{j=1}^{m} a_{ij}\). Compute, for every pair of natural numbers \(n,m \ge 1\):

\[ {\rm a)}~ \sum_{i=1}^{n} \sum_{j=1}^{m} (i+j), \qquad {\rm b)}~ \sum_{i=1}^{n} \sum_{j=1}^{n} i\:j \]

and the value of a) for \(n=3\) and \(m=4\).

Solution

• a) For the inner sum (\(i\) is constant with respect to \(j\)):

  \[ \sum_{j=1}^{m} (i+j) = \sum_{j=1}^{m} i + \sum_{j=1}^{m} j = m\: i + \frac{m^2+m}{2} \]

  then

  \[\begin{align*} \sum_{i=1}^{n} \sum_{j=1}^{m} (i+j) &= \sum_{i=1}^{n} \left( m\: i + \frac{m^2+m}{2}\right) = m \: \frac{n^2+n}{2} + n \: \frac{m^2+m}{2} \\[2ex] &= \frac{n\:m\:(n+1) + n\:m\:(m+1)}{2} = \frac{n\:m\:(n+m+2)}{2} \end{align*}\]

  For \(n=3\) and \(m=4\) we get \(\frac{3 \cdot 4 \cdot 9}{2} = 54\).

• b) Using twice the product by a constant:

  \[ \sum_{i=1}^{n} \sum_{j=1}^{n} i\:j = \sum_{i=1}^{n} \left( i \: \sum_{j=1}^{n} j \right) = \left(\sum_{j=1}^{n} j \right) \: \sum_{i=1}^{n} i = \left( \frac{n^2+n}{2} \right)^2 \]

Exercise 12

Compute, for every natural number \(n \ge 1\), the following double sums in which the upper index of the inner sum depends on \(i\):

\[ {\rm a)}~ \sum_{i=1}^{n} \sum_{j=1}^{i} 1, \qquad {\rm b)}~ \sum_{i=1}^{n} \sum_{j=1}^{i} j \]

and the value of b) for \(n=10\).

Solution

• a) Since \(\sum_{j=1}^{i} 1 = i\), we have:

  \[ \sum_{i=1}^{n} \sum_{j=1}^{i} 1 = \sum_{i=1}^{n} i = \frac{n^2+n}{2} \]

• b) Since \(\sum_{j=1}^{i} j = \frac{i^2+i}{2}\), using formula \(\eqref{EXSQ}\) we have:

  \[\begin{align*} \sum_{i=1}^{n} \sum_{j=1}^{i} j &= \sum_{i=1}^{n} \frac{i^2+i}{2} = \frac{1}{2} \left( \sum_{i=1}^{n} i^2 + \sum_{i=1}^{n} i \right) = \frac{1}{2} \left( \frac{2n^3 + 3n^2 + n}{6} + \frac{n^2+n}{2} \right) \\[2ex] &= \frac{2n^3 + 6n^2 + 4n}{12} = \frac{n\:(n+1)\:(n+2)}{6} \end{align*}\]

  For \(n=10\) we get \(\frac{10 \cdot 11 \cdot 12}{6} = 220\).