Since the first 4 terms of row 12 consists of $1$, $12$, $66$ and $220$

The first 3 terms of row 13, should be

$1+12=13$

$12+66=78$

$220+66=286$.

$(a+b)^n$

Perform the following steps

(1) Based on the exponent $n$ locate the $n^{th}$ row in the triangle. The coefficients of the expansion correspond to the terms in the $n^{th}$ row.

(2) The exponent of $a$ starts from $n$ and decreases by 1 each consecutive terms until it becomes zero on the last term. The exponent of $b$ is zero on the first term and increases by 1 each consecutive term until it reaches $n$.

In this case, $n=5$, from the Pascal’s triangle, the terms at the $5th$ row are $1,5,10,10,5,1$.

It might help you to construct a table showing the coefficient and the corresponding exponent of terms. Then just multiply the terms from the first 3 columns to get the term.

Therefore\\
$(x+2)^5=x^5+10x^4+40x^3+80x^2+80x+32$

for $n=6$, the coefficients are $1,;6,;15,;20,;15,;6,;1$

for $n=3$, the coefficients are $1,3,3,1$

Therefore,

$(2x-3)^3=(2x)^3(-3)^0+(2x)^2(-3)^1+(2x)^1(-3)^2+(2x)^0(-3)^3$

$$
=8x^3-36x^2+54x-27
$$

b.) $(x-1)^6=x^6-6x^5+15x^4-20x^3+15x^2-6x+1$

c.) $(2x-3)^3=8x^3-36x^2+54x-27$

$(a+b)^n$

Perform the following steps

(1) From the exponent $n$, locate the $n^{th}$ row in the Pascal triangle. The coefficients of the expansion correspond to the terms in the $n^{th}$ row.

(2) The exponent of $a$ starts from $n$ and decreases by 1 each consecutive terms until it becomes zero on the last term. The exponent of $b$ is zero on the first term and increases by 1 each consecutive term until it reaches $n$.

In Pascal’s triangle, each row starts with 1. The second number is equal to the row number (called counting number), and third number is a triangular number given by the formula $S_n=0.5n(n-1)$.

Thus for $n=10$, the first three coefficients are $1,10$ and $0.5(10)(9)=45$.

Therefore,

$(x+5)^{10}=x^{10}cdot 5^0 + 10x^9cdot 5^1+45x^8cdot 5^2+…$

$$
implies x^{10}+50x^9+1125x^8+…
$$

Following the method in part (a), the first three coefficients are:

$1, 8,$ and $(0.5)(8)(7)=28$.

Therefore,

$(x-2)^8=1cdot x^8cdot (-2)^0+8cdot x^7cdot (-2)^1+28cdot x^6cdot (-2)^2+….$

$$
implies x^8-16x^7+112x^6+…
$$

Following the method in part (a), the first three coefficients are:

$1, 9,$ and $(0.5)(9)(8)=36$.

Therefore,

$(2x-7)^9=1cdot (2x)^9cdot (-7)^0+9cdot (2x)^8cdot (-7)^1+36cdot (2x)^7cdot (-7)^2+….$

$$
implies 512x^9-16;128x^8+225;729x^7+…
$$

b.) $x^8-16x^7+112x^6+…$

c.) $512x^9-16;128x^8+225;729x^7+…$

$(a+b)^n$

Perform the following steps

(1) Based on the exponent $n$ locate the $n^{th}$ row in the triangle. The coefficients of the expansion correspond to the terms in the $n^{th}$ row.

(2) The exponent of $a$ starts from $n$ and decreases by 1 each consecutive terms until it becomes zero on the last term. The exponent of $b$ is zero on the first term and increases by 1 each consecutive term until it reaches $n$.

In this case, $n=4$, from the Pascal’s triangle, the terms at the $4th$ row are $1,4,6,4,1$.

It might help you to construct a table showing the coefficient and the corresponding exponent of terms. Then just multiply the terms from the first 3 columns to get the term.

Refer to the row 6 of Pascal’s triangle which consists of 1,6,15,20,15,6,1

Therefore

$(y-5)^6$

$=1cdot y^6cdot (-5)^0\+6cdot y^5 cdot (-5)^1\+15cdot y^4cdot (-5)^2\+20cdot y^3cdot (-5)^3\+15cdot y^2cdot (-5)^4\+6cdot y^1 cdot (-5)^5\+1cdot y^0 cdot (-5)^6$

$$
=y^6-30y^5+375y^4-2500y^3+9375y^2-18750y+15625
$$

At row 4, the coefficients are $1,4,6,4,1$

$(3q-4)^4$

$=1cdot(3q)^4cdot(-4)^0\
+4cdot(3q)^3cdot(-4)^1\
+6cdot(3q)^2cdot(-4)^2\
+4cdot(3q)^1cdot(-4)^3\
+1cdot(3q)^0cdot(-4)^4\$\\$=81q^4-432q^3+864q^2-768q+256$

for $n=3$, the coefficients are $1,3,3,1$

$(2x+7y)^3$

$=1cdot (2x)^3cdot(7y)^0\
+3cdot (2x)^2cdot(7y)^1\
+3cdot (2x)^1cdot(7y)^2\
+1cdot (2x)^0cdot(7y)^3\$\\$=8x^3+84x^2y+294xy^2+343y^3$

for $n=6$, the coefficients are $1,6,15,20,15,6,1$

$=1cdot left( sqrt{2} xright)^6 cdot left( sqrt{3} right)^{0}\
+6cdot left( sqrt{2} xright)^5 cdot left( sqrt{3} right)^{1}\
+15cdot left( sqrt{2} xright)^4 cdot left( sqrt{3} right)^{2}\
+20cdot left( sqrt{2} xright)^3 cdot left( sqrt{3} right)^{3}\
+15cdot left( sqrt{2} xright)^2 cdot left( sqrt{3} right)^{4}\
+6cdot left( sqrt{2} xright)^1 cdot left( sqrt{3} right)^{5}\
+1cdot left( sqrt{2} xright)^0 cdot left( sqrt{3} right)^{6}\$\\$=8x^6+24$sqrt{6}$ x^5+180x^4+120$sqrt{6}$ x^3+270x^2+54$sqrt{6}$ x + 27$

for $n=5$, the coefficients are $1,5,10,5,1$

$=1cdot left( 2z^3 right)^5 cdot left( 3y^2 right)^{0}\
+5cdot left( 2z^3 right)^4 cdot left( 3y^2 right)^{1}\
+10cdot left( 2z^3 right)^3 cdot left( 3y^2 right)^{2}\
+10cdot left( 2z^3 right)^2 cdot left( 3y^2 right)^{3}\
+5cdot left( 2z^3right)^1 cdot left( 3y^2 right)^{4}\
+1cdot left( 2z^3 right)^0 cdot left( 3y^2 right)^{5}\$\\$=32z^{15}-240z^{12}y^2+720z^9y^4-1080z^6y^6+810z^3y^8-243y^{10}$

b.) $y^6-30y^5+375y^4-2500y^3+9375y^2-18750y+15625$

c.) $81q^4-432q^3+864q^2-768q+256$

d.) $8x^6+24sqrt{6} x^5+180x^4+120sqrt{6} x^3+270x^2+54sqrt{6} x + 27$

e.) $32z^{15}-240z^{12}y^2+720z^9y^4-1080z^6y^6+810z^3y^8-243y^{10}$

$(a+b)^n$

Perform the following steps

(1) From the exponent $n$, locate the $n^{th}$ row in the Pascal triangle. The coefficients of the expansion correspond to the terms in the $n^{th}$ row.

(2) The exponent of $a$ starts from $n$ and decreases by 1 each consecutive terms until it becomes zero on the last term. The exponent of $b$ is zero on the first term and increases by 1 each consecutive term until it reaches $n$.

In Pascal’s triangle

(1) The first number in each row is 1

(2) The second number is equal to the row number (called counting number)

(3) The third number is a triangular number given by the formula

$S_n=0.5n(n-1)$

When $n=13$, the first three numbers are:

$1, ;13,;$ and $(0.5)(13)(13-1)=78$.

Therefore,

$(x-2)^{13}=1cdot x^{13}cdot (-2)^0+13cdot x^{12}cdot (-2)^1+78x^{11}cdot (-2)^2+…$

$$
(x-2)^{13}=x^{13}-26x^{12}+312x^{11}+…
$$

When $n=9$, the first three numbers are:

$1,;9;,$ and $(0.5)(9)(9-1)=36$

Therefore,

$(3y+5)^9=1cdot(3y)^9cdot (5)^0+9cdot (3y)^8cdot 5^{1}+36cdot (3y)^7cdot 5^2+…$

$$
(3y+5)^9=19683y^9+295245y^8+1;968;300y^7+…
$$

When $n=11$, the first three numbers are:

$1,11,$ and $(0.5)(11)(11-1)=55$

Therefore,

$(z^5-z^3)^{11}=1cdot(z^5)^{11}cdot (-z^3)^0+11cdot (z^5)^{10}cdot (-z^3)^{1}+55cdot (z^5)^9cdot (-z^3)^2+…$

$$
(z^5-z^3)^{11}=z^{55}-11z^{53}+55z^{51}+…
$$

When $n=14$, the first three numbers are:

$1,14,$ and $(0.5)(14)(14-1)=91$

Therefore,

$left(3b^2+dfrac{2}{b}right)^{14}=1cdot (3b^2)^{14}cdot left(dfrac{2}{b}right)^{0}+14cdot (3b^2)^{13}cdot left(dfrac{2}{b}right)^{1}+91cdot (3b^2)^{12}cdot left(dfrac{2}{b}right)^{2}+…$

$$
left(3b^2+dfrac{2}{b}right)^{14}=4;782;969b^{28}-44;641;044b^{25}+193;444;524b^{22}+…
$$

In row 8, the first three numbers in the triangle are:

$1,8,$ and $(0.5)(8)(7)=28$

Therefore,

$left(5x^3+3y^2right)^{8}=1cdot (5x^3)^{8}cdot left(3y^2right)^{0}+8cdot (5x^3)^{7}cdot left(3y^2right)^{1}+91cdot (5x^3)^{6}cdot left( 3y^2 right)^{2}+…$

$$
left(5x^3+3y^2right)^{8}=390;625x^{24}+1;875;000x^{21}y^2+39;375;000x^{18}y^4+…
$$

b.) $(3y+5)^9=19683y^9+295;245y^8+1;968;300y^7+…$

c.) $(z^5-z^3)^{11}=z^{55}-11z^{53}+55z^{51}+…$

d.) $left(3b^2+dfrac{2}{b}right)^{14}=4;782;969b^{28}-44;641;044b^{25}+193;444;524b^{22}+…$

e.) $left(5x^3+3y^2right)^{8}=390;625x^{24}+1;875;000x^{21}y^2+39;375;000x^{18}y^4+…$

$(a+b)^n$

Perform the following steps

(1) From the exponent $n$, locate the $n^{th}$ row in the Pascal triangle. The coefficients of the expansion correspond to the terms in the $n^{th}$ row.

(2) The exponent of $a$ starts from $n$ and decreases by 1 each consecutive terms until it becomes zero on the last term. The exponent of $b$ is zero on the first term and increases by 1 each consecutive term until it reaches $n$.

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

If the $n^{th}$ row consists of the following $t_0,;t_1,;t_2,;t_3,;…t_n$,

$2^n=t_0cdot 1^ncdot 1^{0}+t_1cdot 1^{n-1}cdot 1^1+t_3cdot 1^{n-2}cdot 1^2+…+t_ncdot 1^0cdot 1^n$

$2^n=t_0+t_1+t_2+t_3+…+t_n$

This suggests that the sum of the terms on the $n^{th}$ row of Pascal’s triangle is $2^n$.

For instance,

in row 4: $1+4+6+4+1=2^4$

in row 5: $1+5+10+10+5+1=2^5$

$0^n=(1-1)^n$

Following the notations in part(a),

$0^n=t_0cdot 1^ncdot (-1)^{0}+t_1cdot 1^{n-1}cdot (-1)^1+t_3cdot 1^{n-2}cdot (-1)^2+…+t_ncdot 1^0cdot (-1)^n$

$0^n=t_0-t_1+t_2+…pm t_n$

Observe the alternating signs of the expansion. Also, in any row of the Pascal triangle, we alternate the signs of each term, for example

row 4: $1-4+6-4+1=0$

row 5: $1-5+10-10+5-1=0$

Thus, we have shown by binomial expansion that

$$
0^n=(1-1)^n=0
$$

b.) $0^n=t_0-t_1+t_2+…pm t_n$

All intersections labeled as 1 means there is only one way that Joan can reach that point. For example, the only way to reach the northwest corner is to walk straight 5 N steps.

The intersection labeled as $2$ means there are two ways to reach that point. It could either be (N+E) or (E+N).

For the intersection labeled $3$ on the second column, it could be (N+N+E), (E+N+N) or (N+E+N).

Observe that the number of ways to reach each intersection is the sum of its immediate west and south. Therefore, this appears follow the Pascal triangle.

If we follow the pattern, we’ll find that that the number ways to reach the school is 252.

$_{10}C_{5}=dfrac{10!}{(10-5)!5!}=dfrac{10!}{5!cdot 5!}=252$

$(a+b)^n$

Perform the following steps

(1) From the exponent $n$, locate the $n^{th}$ row in the Pascal triangle. The coefficients of the expansion correspond to the terms in the $n^{th}$ row.

(2) The exponent of $a$ starts from $n$ and decreases by 1 each consecutive terms until it becomes zero on the last term. The exponent of $b$ is zero on the first term and increases by 1 each consecutive term until it reaches $n$.

In Pascal’s triangle, each row starts with 1. The second number is equal to the row number (called counting number), and third number is a triangular number given by the formula

$S_n=0.5n(n-1)$.

$[(x+y)+z]^{10}$

When $n=10$, the first three coefficients are $1,;10,;0.5(10)(10-1)=45$

Therefore,

$[(x+y)+z]^{10}$

$=1cdot(x+y)^{10}cdot z^{0}+10cdot (x+y)^9cdot z^1+45cdot (x+y)^8cdot z^2+…+1cdot (x+y)^0cdot z^{10}$

$$
[(x+y)+z]^{10}=(x+y)^{10}+10(x+y)^9z+45(x+y)^8z^2+…+z^{10}
$$

$(a+b)^n$

Perform the following steps

(1) From the exponent $n$, locate the $n^{th}$ row in the Pascal triangle. The coefficients of the expansion correspond to the terms in the $n^{th}$ row.

(2) The exponent of $a$ starts from $n$ and decreases by 1 each consecutive terms until it becomes zero on the last term. The exponent of $b$ is zero on the first term and increases by 1 each consecutive term until it reaches $n$.

In Pascal’s triangle, each row starts with 1. The second number is equal to the row number (called counting number), and third number is a triangular number given by the formula

$S_n=0.5n(n-1)$.

When $n=6$, the terms in the Pascal’s triangle are $1,;6,;15,;20,;15,;6,;1$

$(3x-5y)^6$

$=1cdot (3x)^6cdot (-5y)^0\
+6cdot (3x)^5cdot (-5y)^1\
+15cdot (3x)^4cdot (-5y)^2\
+20cdot (3x)^3cdot (-5y)^3\
+15cdot (3x)^2cdot (-5y)^4\
+6cdot (3x)^1cdot (-5y)^5\
+1cdot (3x)^0cdot (-5y)^6$

$$
=729x^6-7290x^5y+30;375x^4y^2-67500x^3y^3+84;375 x^2y^2-56250xy^5+15625y^6
$$

$(a+b)^n$

Perform the following steps

(1) From the exponent $n$, locate the $n^{th}$ row in the Pascal triangle. The coefficients of the expansion correspond to the terms in the $n^{th}$ row.

(2) The exponent of $a$ starts from $n$ and decreases by 1 each consecutive terms until it becomes zero on the last term. The exponent of $b$ is zero on the first term and increases by 1 each consecutive term until it reaches $n$.

In Pascal’s triangle, each row starts with 1. The second number is equal to the row number (called counting number), and third number is a triangular number given by the formula

$S_n=0.5n(n-1)$.

$t_n=An^3+Bn^2+Cn+D$

The first difference of any consecutive term $t_{n}$ and $t_{n+1}$ is

$(a+b)^n$

Perform the following steps

(1) From the exponent $n$, locate the $n^{th}$ row in the Pascal triangle. The coefficients of the expansion correspond to the terms in the $n^{th}$ row.

(2) The exponent of $a$ starts from $n$ and decreases by 1 each consecutive terms until it becomes zero on the last term. The exponent of $b$ is zero on the first term and increases by 1 each consecutive term until it reaches $n$.

In Pascal’s triangle, each row starts with 1. The second number is equal to the row number (called counting number), and third number is a triangular number given by the formula

$S_n=0.5n(n-1)$.