1 Summations

1.1 Solution 1

1.1.1 Solution a

From theorem 5.1.1, we can rewrite them as individual summations:

i, n = var('i,n')
a = sum(i^2, i , 1, n)
a

b = sum(-2*i, i, 1, n)
b

c = sum(3, i, 1, n)
c

d = a + b + c
d
d.factor()

1/3*n^3 + 1/2*n^2 + 1/6*n
-n^2 - n
3*n
1/3*n^3 - 1/2*n^2 + 13/6*n
1/6*(2*n^2 - 3*n + 13)*n

We can indeed confirm the result:

i,n = var('i n')
f = i^2 - (2*i) + 3
sum(f(i), i, 1, n).factor()

1/6*(2*n^2 - 3*n + 13)*n

1.1.2 Solution b

sum((i^2 - (2*i) + 3), i, 1, 5)

f
f(1) + f(2) + f(3) + f(4) + f(5)

i^2 - 2*i + 3
40

1.2 Solution 2

From theorem 5.1.1 and results from 1.4, we can rewrite them as individual summations:

i, n = var('i n')
f = i^3 - (4*i)

a = sum(i^3, i, 1, n).factor()
b = sum((-4*i),i,1,n).factor()

(a + b).factor()

1/4*(n^2 + n - 8)*(n + 1)*n

Let's check our solution:

sum(f,i,1,n).factor()

1/4*(n^2 + n - 8)*(n + 1)*n

1.3 Solution 3

From theorem 5.1.1 and results from 1.4, we can rewrite them as individual summations:

j, n = var('j n')

a = sum(2*(j^3), j, 1, n)
b = sum(-1*j, j, 1, n)

(a - b).factor()

1/2*(n^2 + n + 1)*(n + 1)*n

1.4 Solution 4

From theorem 5.1.1 and results from 1.4, we can rewrite them as individual summations:

i, t = var('i t')

a = sum(3 * (i^2), i, 1, t)
b = sum(-i, i, 1, t)
c = sum(2, i, 1, t)

(a + b + c).factor()

(t^2 + t + 2)*t

Let's verify the solution:

f = 3*(i^2) - i + 2
sum(f(i), i, 1, n).factor()

(n^2 + n + 2)*n

1.5 Solution 5

Let's expand \((1+1)^3\):

i = var('i')
((i+1)^3).expand()

i^3 + 3*i^2 + 3*i + 1

Now from theorem 5.1.1 and results from 1.4, we can rewrite them as individual summations:

i = var(i)

a = sum(i^3, i, 1, n)
b = sum(3*(i^2), i, 1, n)
c = sum(3*i, i, 1, n)
d = sum(1, i, 1, n)

(a+b+c+d).factor()

1/4*(n^2 + 3*n + 4)*(n + 3)*n

1.6 Solution 6

Let's find the solution from \(i = 1\) to \(i = n+3\) in this case:

i, n = var('i n')
whole = sum(i^2, i, 1, n+3).factor()
whole

1/6*(2*n + 7)*(n + 4)*(n + 3)

Now let's compute the summation from \(i = 1\) to \(i = 4\):

till_4 = sum(i^2, i, 1, 4)
till_4

The final solution is:

(whole - till_4).factor()

1/6*(2*n^2 + 23*n + 96)*(n - 1)

Let's verify the solution:

i, n = var('i n')
sum(i^2, i, 5, (n+3)).factor()

1/6*(2*n^2 + 23*n + 96)*(n - 1)

1.7 Solution 7

1.7.1 Solution a

Now from theorem 5.1.1 and results from 1.4, we can rewrite them as individual summations:

i, n = var('i n')

a = sum(i^2, i, 1, n)
b = sum(5*i, i, 1, n)
(a+b).factor()

1/3*(n + 8)*(n + 1)*n

1.7.2 Solution b

Now from theorem 5.1.1 and results from 1.4, we can rewrite them as individual summations. Let's find the sum from \(j = 1\) to \(n+5\)

j, n = var('j n')

a = sum(j^2, j, 1, n+5)
b = sum(-5*j, j, 1, n+5)
(a + b).factor()

1/3*(n + 6)*(n + 5)*(n - 2)

Now let's compute the same sum from \(j = 1\) to \(5\):

a1 = sum(j^2, j, 1, 5)
b1 = sum(-5*j, j, 1, 5)
(a + b - a1 - b1).factor()

1/3*(n + 8)*(n + 1)*n

1.7.3 Solution c

We have already shown from (a) and (b) that they are equal. Let's expand them:

i = var(i)
f = i^2 + 5*i
[f(1) , f(2) , f(3) , f(n).factor()]

g = j^2 - 5*j
[g(6), g(7), g(8), g(n+6).factor()]

[6, 14, 24, (n + 5)*n]
[6, 14, 24, (n + 6)*(n + 1)]

We can see how each of them expand to the same value when started from their staring interval. That's why both come to the same value:

f(n).factor()
g(n+5).factor()

(n + 5)*n
(n + 5)*n

1.8 Solution 8

1.8.1 Part one

We need to prove this:

\(\sum_{i=1}^n (ca_i) = ca_1 + ca_2 + ... + ca_n = c(a_1 + a_2 + ... + a_n) = c\sum_{i=1}^n a_i\)

We will prove using mathematical induction.

Base case: \(n= 1\)

\(\sum_{i=1}^n ca_i = ca_1\)

Induction step: Suppose \(\sum_{i=1}^n (ca_i) = c\sum_{i=1}^n a_i\)

\(\sum_{i=1}^{n+1} (ca_i) = \sum_{i=1}^n (ca_i) + ca_{n+1}\)

\(= c\sum_{i=1}^n a_i + ca_{n+1}\) (Induction case)

\(= c(\sum_{i=1}^n a_i + a_{n+1})\)

\(= c(\sum_{i=1}^{n+1} a_i)\)

This completes the mathematical induction proof.

1.8.2 Part two

We need to prove this:

\(\sum_{i=1}^n (a_i + b_i) = \sum_{i=1}^n a_i + \sum_{i=1}^n b_i\)

We will prove using mathematical induction.

Base case: \(n = 1\)

\(\sum_{i=1}^n (a_i + b_i) = a_i + b_i\)

Induction step: Suppose \(\sum_{i=1}^n (a_i + b_i) = \sum_{i=1}^n a_i + \sum_{i=1}^n b_i\)

\(\sum_{i=1}^{n+1} (a_i + b_i)\)

\(= \sum_{i=1}^{n} (a_i + b_i) + a_{n+1} + b_{n+1}\)

\(= \sum_{i=1}^n a_i + \sum_{i=1}^n b_i + a_{n+1} + b_{n+1}\)

\(= \sum_{i=1}^n a_{i+1} + \sum_{i=1}^n b_{i+1}\)

\(= \sum_{i=1}^{n+1} a_{i} + \sum_{i=1}^{n+1} b_{i}\)

This completes the mathematical induction proof.

1.9 Solution 9

\(\sum_{i=1}^n [(i+1)^4 - i^4]\)

We need to derive equation 5.9 using the above equation.

Let's first expand the above equation:

  i = var('i')
  f1 = (i+1)^4
  f2 = -i^4
  [f1(1), f2(1), f1(2), f2(2), f1(3), f2(3), f1(4), f2(4), f1(i-1), f2(i-1), f1(i), f2(i)]

[16, -1, 81, -16, 256, -81, 625, -256, i^4, -(i - 1)^4, (i + 1)^4, -i^4]

You can see that the first term is cancelled by the fourth term. The third term is cancelled by two terms following it. So apart from \(-1\) and \((i+1)^4\) everything else is cancelled.

So, \(\sum_{i=1}^n [(i+1)^4 - i^4] = (i+1)^4 - 1\)

We also know that:

x = (i+1)^4 - i^4
x.simplify_full()

4*i^3 + 6*i^2 + 4*i + 1

So, \(\sum_{i=1}^n [(i+1)^4 - i^4] = \sum_{i=1}^n(4*i^3 + 6*i^2 + 4i + 1)\)

Combining both the equations:

\((i+1)^4 - 1 = 4. \sum_{i=1}^n i^3 + \sum{i=1}^n(6*i^2 + 4*i + 1)\)

sum(((6*i^2) + (4*i) + 1), i, 1, n)

2*n^3 + 5*n^2 + 4*n

Reducing it further:

\((n+1)^4 - 1 = 4. \sum_{i=1}^n i^3 + 2*n^3 + 5*n^2 + 4n\)

\(4\sum_{i=1}^n i^3 = (n+1)^4 - 1 - 2*n^3 - 5*n^2 - 4n\)

Reducing the equations:

n = var('n')
((n+1)^4 - 1 - (2*n^3) - (5*n^2) - (4*n)).factor()

(n + 1)^2*n^2

So, \(\sum_{i=1}^n i^3 = \dfrac{n^2(n+1)^2}{4}\)

1.10 Solution 10

We need to find alternative derivation of equation 5.4 by evaulating \(\sum_{i=1}^n [(i+1)^2-i^2]\)

Let's first expand the equation:

i = var(i)
f1 = (i+1)^2
f2 = -i^2

[f1(1), f2(1), f1(2), f2(2), f1(3), f2(3), f1(i-1), f2(i-1), f1(i), f2(i)]

[4, -1, 9, -4, 16, -9, i^2, -(i - 1)^2, (i + 1)^2, -i^2]

Apart from \(-1\) and \((i+1)^2\) everything else gets cancelled out.

So, \(\sum_{i=1}^n [(i+1)^2-i^2] = (n+1)^2 - 1\)

We also know that:

((i+1)^2 - i^2).simplify_full()

2*i + 1

So, \(\sum_{i=1}^n [(i+1)^2-i^2] = \sum_{i=1}^n 2i + 1\)

Combining both the equations:

\((n+1)^2 - 1 = \sum_{i=1}^n 2i + 1\)

\(\sum_{i=1}^n i = \dfrac{(n+1)^2 - n}{2}\)

((n+1)^2 - n - 1).simplify_full()

n^2 + n

So, \(\sum_{i=1}^n i = \dfrac{n^2 + n}{2}\)

1.11 Solution 11

We need to prove \(\sum_{i=1}^n i = \dfrac{n^2 + n}{2}\)

We will prove using mathematical induction.

Base case: \(n = 1\)

\(\sum_{i=1}^n i = \dfrac{1 + 1}{2} = \dfrac{1}{1} = 1\)

\(\sum_{i=1}^1 = 1\)

The base case holds.

Induction step: Suppose \(\sum_{i=1}^n i = \dfrac{n^2 + n}{2}\).

\(\sum_{i=1}^{n+1} i = \sum_{i=1}^n i + (n+1)\)

\(\sum_{i=1}^{n+1} i = \dfrac{n^2 + n}{2} + (n+1)\) (Induction step)

(((n^2 + n)/2) + (n+1)).factor()

1/2*(n + 2)*(n + 1)

((n+1)*(n+2)).simplify_full()

n^2 + 3*n + 2

So, \(\sum_{i=1}^{n+1} i = \dfrac{(n+1)^2 + (1+n)}{2}\)

This completes the mathematical induction proof.

1.12 Solution 12

1.12.1 Solution a

\(\sum_{i=1}^n \dfrac{1}{i(i+1)}\)

i, n = var('i n')
f = (1/(i*(i+1)))
[sum(f(i), i, 1, 1), sum(f(i), i, 1, 2), sum(f(i), i, 1, 3), sum(f(i), i, 1, 4)]

[1/2, 2/3, 3/4, 4/5]

So, I'm guessing the formula for the sum as:

\(\dfrac{n}{n+1}\)

Let's try to prove it using mathematical induction.

Base case: \(n = 1\)

\(\sum_{i=1}^n \dfrac{1}{i(i+1)} = \dfrac{1}{1(1+1)} = \dfrac{1}{2}\)

\(\dfrac{n}{n+1} = \dfrac{1}{2}\)

The base case holds.

Induction step: Suppose \(\sum_{i=1}^n \dfrac{1}{i(i+1)} = \dfrac{n}{n+1}\)

\(\sum_{i=1}^{n+1} \dfrac{1}{i(i+1)} = \sum_{i=1}^n \dfrac{1}{i(i+1)} + \dfrac{1}{(n+1)(n+2)}\)

\(= \dfrac{n}{n+1} + \dfrac{1}{(n+1)(n+2)}\) (Induction step)

\(= \dfrac{1}{n+1}(n + \dfrac{1}{n+2})\)

\(= \dfrac{1}{n+1} \dfrac{n^2 + 2n + 1}{n+2}\)

\(= \dfrac{1}{n+1} \dfrac{(n+1)^2}{n+2}\)

\(= \dfrac{n+1}{n+2}\)

So, \(\sum_{i=1}^{n+1} \dfrac{1}{i(i+1)} = \dfrac{n+1}{n+2}\)

This completes the mathematical induction proof.

We can also verify this using SageMath that our guessed formula is right:

sum(f(i),i,1,n)

n/(n + 1)

1.12.2 Solution b

We need to verify that \(\dfrac{1}{i+1} = \dfrac{1}{i} - \dfrac{1}{i+1}\)

\(\dfrac{1}{i} - \dfrac{1}{i+1}\)

\(= \dfrac{i+1-1}{i(i+1)}\)

\(= \dfrac{i}{i(i+1)}\)

\(= \dfrac{1}{i+1}\)

Now let's use this fact to write part(a) as telescoping sum.

i = var(i)
f1 = (1/i)
f2 =  - (1/(i+1))
[f1(1), f2(1), f1(2), f2(2), f1(3), f2(3), f1(n-1), f2(n-1), f1(n), f2(n)]

[1, -1/2, 1/2, -1/3, 1/3, -1/4, 1/(n - 1), -1/n, 1/n, -1/(n + 1)]

We can see how all the terms cancel out except the first one and the last term. So,

\(\sum_{i=1}^{n} \dfrac{1}{i(i+1)} = 1 - \dfrac{1}{n+1}\)

\(= \dfrac{n}{n+1}\)

1.13 Solution 13

Suppose \(r\) is a real number and \(r \neq 1\)

1.13.1 Solution a

\(\sum_{i=1}^n [r^{i+1} - r^i]\)

i,r = var('i r')
f1 = r^(i+1)
f2 = -r^i

[f1(r=r,i=1), f2(r=r,i=1), f1(r=r,i=2), f2(r=r,i=2), f1(r=r,i=3), f2(r=r,i=3), f1(r=r,i=n-1), f2(r=r,i=n-1), f1(r=r,i=n), f2(r=r,i=n)]

[r^2, -r, r^3, -r^2, r^4, -r^3, r^n, -r^(n - 1), r^(n + 1), -r^n]

Some observations:

First term is cancelled with the fourth term
Third term is cancelled with the sixth term
Second term is not cancelled
Second term from the last is not cancelled

So, \(\sum_{i=1}^n [r^{i+1} - r^i] = r^{n+1} - r\)

Let's verify the solution using SageMath:

sum(r^(i+1) - r^i, i, 1, n)

r^(n + 1) - r

1.13.2 Solution b

\(\sum_{i=1}^n [(r-1)r^i] = (r-1)\sum_{i=1}^n r^i\)

We need to find the formula for \(\sum_{i=1}^n r^i\)

We know from (a) that \(\sum_{i=1}^n [r^{i+1} - r^i] = r^{n+1} - r\).

So, \(r^{n+1} - r = (r-1)\sum_{i=1}^n r^i\)

\(\sum_{i=1}^n r^i = \dfrac{r^{n+1} - r}{r-1}\)

1.13.3 Solution c

\(\sum_{i=1}^n (\dfrac{1}{2})^i\)

Using the formula from (b),

r, n = var('r n')
f = (r^(n+1) - r)/(r - 1)
f(r=(1/2)).simplify()

-1/2^n + 1

1.14 Solution 14

\(\sum_{i=1}^n i2^i = 2 + (n-1)2^{n+1}\)

Let's prove it using mathematical induction.

Base case: \(n = 1\)

\(\sum_{i=1}^n i2^i = 1.2^1 = 2\)

\(2 + (n-1)2^{n+1} = 2 + 0 = 2\)

The base case holds.

Induction case:

Suppose \(\sum_{i=1}^n i2^i = 2 + (n-1)2^{n+1}\).

\(\sum_{i=1}^{n+1} i2^i = \sum_{i=1}^n i2^i + (n+1)2^{n+1}\)

\(= 2 + (n-1)2^{n+1} + (n+1)2^{n+1}\) (Induction step)

\(= 2 + 2^{n+1}(n-1+n+1)\)

\(= 2 + 2^{n+1}(2n)\)

\(= 2 + n2^{n+2}\)

So, \(\sum_{i=1}^{n+1} i2^i = 2 + n2^{n+2}\)

This completes the mathematical induction proof.

1.15 Solution 15

Suppose \(\sin (\dfrac{\theta}{2}) \neq 0\) We need to prove that

\(\sum_{i=1}^n \cos(i\theta) = \dfrac{\cos(\dfrac{(n+1)\theta}{2}) \sin(\dfrac{n\theta}{2})}{\sin(\theta / 2)}\)

Let's prove it using mathematical induction.

Base case: \(n = 1\)

\(\sum_{i=1}^n \cos(i\theta) = cos (\theta)\)

\(\dfrac{\cos(\dfrac{(n+1)\theta}{2}) \sin(\dfrac{n\theta}{2})}{\sin(\theta / 2)} = \dfrac{\cos(\theta) \sin(\theta / 2)}{\sin(\theta / 2)} = \cos(\theta)\)

The base case holds.

Inducation case:

Suppose \(\sum_{i=1}^n \cos(i\theta) = \dfrac{\cos(\dfrac{(n+1)\theta}{2}) \sin(\dfrac{n\theta}{2})}{\sin(\theta / 2)}\)

\(\sum_{i=1}^{n+1} \cos(i\theta) = \sum_{i=1}^{n} \cos(i\theta) + cos((n+1) \theta)\)

\(= \dfrac{\cos(\dfrac{(n+1)\theta}{2}) \sin(\dfrac{n\theta}{2})}{\sin(\theta / 2)} + cos((n+1) \theta)\)

We know that \(\cos u \sin v = \dfrac{1}{2}[\sin(u+v)-\sin(u-v)]\)

\(\cos(\dfrac{(n+1)\theta}{2}) \sin(\dfrac{n\theta}{2}) = \dfrac{1}{2}[\sin(\dfrac{(n+1)\theta}{2} + \dfrac{n\theta}{2}) - \sin(\dfrac{(n+1)\theta}{2} - \dfrac{n\theta}{2})]\)

\(= \dfrac{1}{2}[\sin(\dfrac{(2n + 1)\theta}{2}) - \sin(\dfrac{\theta}{2})]\)

Dividing the above equation by \(\sin (\theta / 2)\)

\(= \dfrac{1}{2}\dfrac{\sin(\dfrac{(2n + 1)\theta}{2})}{\sin(\dfrac{\theta}{2})} - \dfrac{1}{2}\)

Combining it with the orignal equation:

\(= \dfrac{1}{2}\dfrac{\sin(\dfrac{(2n + 1)\theta}{2})}{\sin(\dfrac{\theta}{2})} - \dfrac{1}{2} + \cos((n+1) \theta)\)

n, x = var('n x')

a = sin(((2*n + 1)*x)/2)
b = sin(x/2)
c = 1/2 * (a/b)
assume(b > 0)
f = c - (1/2) + cos((n + 1)*x)

f.simplify_full()

1/2*(2*cos((n + 1)*x)*sin(1/2*x) + sin(1/2*(2*n + 1)*x) - sin(1/2*x))/sin(1/2*x)

(todo: Fix this proof)