Series

∑ stands for 'sum of'

$\sum_{r=1}^{3} r = 1+2+3 = 6$

e.g.

$\sum_{r=1}^{3} r = 1+2+3 = 6$

$\sum_{r=3}^{5} r^2 = 3^2 + 4^2 + 5^2$

$\sum_{r=3}^{8} (2x-1) = 5+7+9+11+13+15$

Discuss the following and find the sum if $n = 8$

a) $\sum_{n=1}^{\infty} \frac{n+4}{(n+2)(n+2)}$

b) $\sum_{n=1}^{\infty} \frac{1}{n\sqrt{n^2 + 1}}$

I. If $a_n = n^2 + 3n + 1$ determine an expression for $n$

II. If $a_n = n^3 + 2n^2 + 4n$ evaluate

a) $a_1$ b) $a_4$ c)

Proof:

1. Evaluate

Prove that $n(n^2 + 5)$ is exactly divisible by 3 for all positive integers $n$

Proof: I

Let $n = 1$; $1(1^2 + 5) = 6 = 3 \times 2$

$n = 2$; $2(2^2 + 5) = 18 = 3 \times 6$

$n = 3$; $3(3^2 + 5) = 42 = 3 \times 14$

$n = 4$; $4(4^2 + 5) = 84 = 3 \times 28$

$n = 7$; $7(7^2 + 5) = 378 = 3 \times 126$

Proof: II

i) Let $n = 1 = 1(1^2 + 5) = 6 = 3 \times 2$

ii) Let $n(n^2 + 5)$ be divisible for $n = k$

i.e. $k(k^2 + 5) = 3p$, where $p$ is any integers

iii) When $n = k + 1$

$(k + 1)((k + 1)^2 + 5) = (k + 1)(k^2 + 2k + 1 + 5)$

$= (k + 1)((k^2 + 5) + (2k + 1))$

$= k(k^2 + 5) + k(2k + 1) + (k^2 + 5) + (2k + 1)$

$= 3p + 2k^2 + k + k^2 + 5 + 2k + 1$

$= 3p + 3k^2 + 3k + 6$

$= 3(p + k^2 + k + 2)$

Since $p$ and $k$ are positive integers

So the number in the bracket is positive

iv) Since when $n = 1$ the values $1(1^2 + 5)$ is divisible by 3 then the value $n(n^2 + 5)$ will be divisible by 3 for $n = 2$, $n = 3$, $n = 4$…… by the above working

→ $n(n^2 + 5)$ is divisible by 3 for all $n \in$ +

It states if $s_1, s_2, s_3 \ldots S_n \ldots$ is a sequence of statements and if

i) $s_1$ is true

ii) $S_n \rightarrow S_{n+1}$, $n = 1, 2, 3 \ldots$ are true, then $s_1, s_2, s_3 \ldots S_n \ldots$ are true statement

1. Prove by mathematical induction that $2 + 4 + 6 + \ldots + 2n = n(n + 1)$

Solution

When $n = 1$

L. H. S = 2, R. H. S = $1(1 + 1) = 2$

L. H.S = R. H. S

It is true for $n = 1$

Let the statement be true for $n = k$

$2 + 4 + 6 + \ldots + 2k = k(k + 1)$

Required to prove when $n = k + 1$

$2 + 4 + 6 + \ldots + 2k + 2(k + 1) = k(k + 1) + 2(k + 1)$

$= k^2 + k + 2k + 2$

$= k^2 + 3k + 2$

$= k^2 + k + 2k + 2$

$= k(k + 1) + 2(k + 1)$

$= (k + 1)(k + 2)$

Which is the same as putting $n = k + 1$ in the formula

Since $n = 1$ gave a true statement, $n = 2$, $n = 3$, $n = 4 \ldots$ will be true statement as worked above

2. Prove by induction that

Solution

Proof:

When $n = 1$,

Also $n = 1$ give

L.H.S = R. H. S

Let the statement be true for $n = k$

Let

Required to prove when $n = k + 1$:

Which is the same as putting $n = k + 1$ in the form

Since $n = 1$ gave a true statement

$n = 2$, $n = 3$, $n = 4 \ldots$ will give true statement

3. Prove that

Solution

Proof:

When $n = 1$

L.H. S = $3 \times 1 - 2 = 1$

R.H.S =

L.H.S = R.H.S

Consider $a^{\frac{1}{2}} \times a^{\frac{1}{2}} = a^1$

Similarly

In general

$a^{\frac{m}{n}} = \sqrt[n]{a^m}$

Consider $a^m \times a^0 = a^{m + 0} \rightarrow a^0 = 1$ ∵ $a^0 = 1$