4. Formulae and Powers to remember

4.1 Algebraic Expansions

The binomial expansion formula is provided below.

$(a + b)^n =$ $^n C_0 a^n b^0 +$ $^n C_1 a^{n-1} b^1 +$ $^n C_2 a^{n-2} b^2 + ... +$ $^n C_n a^{0} b^n$

Note: In the formula above, $^nC_{r} = \dfrac{n!}{r! \space (n - r)!}$. This is detailed in Permutations and Combinations lesson.

It is easy to remember this if you observe the following.

1) The first term is $^nC_0a^nb^0$, where $r = 0$.
2) For every subsequent term, $r$ increases by $1$, power of $a$ reduces by $1$ and power of $b$ increases by $1$.
3) $\therefore$ the last term is $^nC_na^0b^n$
4) Note that if the power is $n$, there will be $n + 1$ terms.

For instance, to derive the expansion for $(a+b)^2$, we substitute $n=2$ in the formula above and get the following.

$(a + b)^2 =$ $^2 C_0 a^2 b^0 +$ $^2 C_1 a^1 b^1 +$ $^2 C_2 a^0 b^2$

⇒ $ (a + b)^2 = a^2 + 2ab + b^2$

Algebraic Formulae to Memorise

Expression	Expansion
$(a+b)^{n}$	$^n C_0 a^n b^0 +$ $^n C_1 a^{n-1} b^1 + ... +$ $^nC_n a^{0} b^n$
$a^{n} - b^{n}$	$(a - b) (a^{n - 1} + a^{n - 2} b + ... + b^{n - 1})$
$(a+b)^2$	$a^2 + 2 ab + b^2$
$(a-b)^2$	$a^2 - 2 ab + b^2$
$a^2-b^2$	$(a + b) (a - b)$
$a^2+b^2$	$(a + b)^2 - 2 ab$; or $(a - b)^2 + 2 ab$
$(a+b)^3$	$a^3 + 3ab (a + b) + b^3$; or $a^3 + 3 a^2 b + 3 a b^2 + b^3$
$(a-b)^3$	$a^3 - 3ab (a - b) - b^3$; or $a^3 - 3 a^2 b + 3 a b^2 - b^3$
$a^3+b^3$	$(a + b) (a^2 - ab + b^2) $; or $(a + b)^3 - 3 ab (a + b)$
$a^3-b^3$	$(a - b) (a^2 + ab + b^2) $; or $(a - b)^3 + 3 ab (a - b)$
$(a + b + c)^2$	$a^2 + b^2 + c^2 + 2 ab + 2 bc + 2 ca$
$a^3 + b^3 + c^3 - 3 abc$	$(a + b + c) (a^2 + b^2 + c^2 - ab - bc - ca)$
If $a+b+c=0$, then $a^3 + b^3 + c^3$ $=$	$3 abc$

The following are to be noted as well.
1) $(a^{n} + b^{n})$ is divisible by $(a + b)$ if $n$ is odd.
2) $(a^{n} - b^{n})$ is divisible by $(a + b)$ if $n$ is even.

Example 13

The remainder, when $(15^{23} + 23^{23})$ is divided by $19$, is:
[CAT 2004]

(1) $4$ (2) $15$ (3) $0$ (4) $18$

Solution

$(a^{n} + b^{n})$ is divisible by $(a + b)$ if $n$ is odd.

$\therefore (15^{23} + 23^{23})$ is divisible by $15 + 23 = 38$

$\therefore (15^{23} + 23^{23})$ will perfectly divide $19$ and leave a remainder of $0$.

Answer: (3) $0$

Example 14

If R = $\dfrac{30^{65} - 29^{65}}{30^{64} + 29^{64}}$, then:
[CAT 2005]

(1) $0$ < R < $0.1$ (2) $0.1$ < R < $0.5$ (3) $0.5$ < R < $1.0$ (4) R > $1.0$

Solution

$a^{n} - b^{n} = (a - b) (a^{n - 1} + a^{n - 2} b + ... + b^{n - 1})$

$30^{65} - 29^{65} = (30 - 29) (30^{64} + 30^{63} \times 29 + ... + 30 \times 29^{63} + 29^{64})$

= $(30^{64} + 29^{64}) + (30^{63} \times 29 + ... + 30 \times 29^{63})$

$\therefore$ R = $\dfrac{(30^{64} + 29^{64}) + (30^{63} \times 29 + ... + 30 \times 29^{63})}{30^{64} + 29^{64}}$

$= 1 + \dfrac{(30^{63} \times 29 + ... + 30 \times 29^{63})}{30^{64} + 29^{64}} \gt 1$

Answer: (4) R > $1.0$

Additionally, to improve your speed in answering questions, it is best if you memorise the following
(1) Multiplication Tables from $1$ to $15$; and
(2) Squares and higher powers listed below provided in $4.2$ and $4.3$.

4.2 Squares

$1^2 = 1$ $9^2 = 81$ $17^2 = 289$ $25^2 = 625$

$2^2 = 4$ $10^2 = 100$ $18^2 = 324$ $26^2 = 676$

$3^2 = 9$ $11^2 = 121$ $19^2 = 361$ $27^2 = 729$

$4^2 = 16$ $12^2 = 144$ $20^2 = 400$ $28^2 = 784$

$5^2 = 25$ $13^2 = 169$ $21^2 = 441$ $29^2 = 841$

$6^2 = 36$ $14^2 = 196$ $22^2 = 484$ $30^2 = 900$

$7^2 = 49$ $15^2 = 225$ $23^2 = 529$ $31^2 = 961$

$8^2 = 64$ $16^2 = 256$ $24^2 = 576$ $32^2 = 1024$

4.3 Higher Powers

$2^1 = 2$	$2^2 = 4$	$2^3 = 8$	$2^4 = 16$
$2^5 = 32$	$2^6 = 64$	$2^7 = 128$	$2^8 = 256$
$2^9 = 512$	$2^{10} = 1024$	$2^{11} = 2048$	$2^{12} = 4096$

$3^1 = 3$	$3^2 = 9$	$3^3 = 27$	$3^4 = 81$
$3^5 = 243$	$3^6 = 729$	$3^7 = 2187$	$3^8 = 6561$

$4^1 = 4$	$4^2 = 16$	$4^3 = 64$	$4^4 = 256$
$4^5 = 1024$	$4^6 = 4096$	$$	$$

$5^1 = 5$	$5^2 = 25$	$5^3 = 125$	$5^4 = 625$
$5^5 = 3125$	$5^6 = 15625$	$$	$$

$6^1 = 6$	$6^2 = 36$	$6^3 = 216$	$6^4 = 1296$

$7^1 = 7$	$7^2 = 49$	$7^3 = 343$	$7^4 = 2401$

$8^1 = 8$	$8^2 = 64$	$8^3 = 512$	$8^4 = 4096$

$9^1 = 9$	$9^2 = 81$	$9^3 = 729$	$9^4 = 6561$

$11^1 = 11$	$11^2 = 121$	$11^3 = 1331$	$11^4 = 14641$

$12^1 = 12$	$12^2 = 144$	$12^3 = 1728$	$$