diff --git a/src/content/questions/comp2804/2014-fall-midterm/7/solution.md b/src/content/questions/comp2804/2014-fall-midterm/7/solution.md index 42878a37..31fcbe69 100644 --- a/src/content/questions/comp2804/2014-fall-midterm/7/solution.md +++ b/src/content/questions/comp2804/2014-fall-midterm/7/solution.md @@ -1,11 +1,11 @@ ${(5x-36)}^{100}$ -$=\sum^{100}_{k=0} \binom{100}{k}{(5x)}^k {(-3y)}^{n-k}$ +$ = \sum^{100}_{k=0} \binom{100}{k} {(5x)}^{n-k} {(-3y)}^{k} $ -$=\binom{100}{20}{(5x)}^{20} {(-3y)}^{80}$ +We only consider $k=80$, as it results in $y^{80}$. -$=\binom{100}{20}5^{20} 3^{80} x^{20} y^{80}$ +$ = \binom{100}{80} \cdot {(5x)}^{100-80} \cdot {(-3y)}^{80} $ -$=\binom{100}{80}5^{20} 3^{80} x^{20} y^{80}$ +$ = \binom{100}{80} \cdot 5^{20} \cdot {(-3)}^{80} \cdot x^{20} \cdot y^{80} $ -$=\binom{100}{80}5^{20} 3^{80}$ (this is the coefficient) +$ = \binom{100}{80} \cdot 5^{20} \cdot 3^{80} $ (final answer, i.e. the coefficient of $x^{20} y^{80}$) diff --git a/src/content/questions/comp2804/2015-fall-final/4/solution.md b/src/content/questions/comp2804/2015-fall-final/4/solution.md index 3a0f9a40..fb663a23 100644 --- a/src/content/questions/comp2804/2015-fall-final/4/solution.md +++ b/src/content/questions/comp2804/2015-fall-final/4/solution.md @@ -1,7 +1,11 @@ -$ = \sum^{20}\_{k=0} \binom{20}{k} {(-3x)}^{k} {(5y)}^{20-k} $ +$ = \sum^{20}_{k=0} \binom{20}{k} {(-3x)}^{n-k} {(5y)}^{k} $ -$ = \binom{20}{15} {(-3)}^{15} {5}^{5} x^{15} y^5 $ +We only consider $k=5$, as it results in $y^{5}$. -$ = - \binom{20}{15} {(3)}^{15} {5}^{5} x^{15} y^5 $ +$ = \binom{20}{5} \cdot {(-3x)}^{20-5} \cdot {(5y)}^{5} $ -Thus, the coefficient of $x^{15}y^{5}$ in the expansion of ${(-3x + 5y)}^{20}$ is $ - \binom{20}{15} {(3)}^{15} {5}^{5} $ +$ = \binom{20}{5} \cdot {(-3)}^{15} \cdot {5}^{5} \cdot x^{15} \cdot y^5 $ + +$ = - \binom{20}{5} \cdot {3}^{15} \cdot {5}^{5} \cdot x^{15} \cdot y^5 $ + +Thus, the coefficient of $x^{15}y^{5}$ in the expansion of ${(-3x + 5y)}^{20}$ is $ - \binom{20}{5} \cdot {3}^{15} \cdot {5}^{5} $ diff --git a/src/content/questions/comp2804/2015-fall-midterm/7/index.md b/src/content/questions/comp2804/2015-fall-midterm/7/index.md index cba52a19..cf0cef72 100644 --- a/src/content/questions/comp2804/2015-fall-midterm/7/index.md +++ b/src/content/questions/comp2804/2015-fall-midterm/7/index.md @@ -8,5 +8,5 @@ solution: comp2804/2015-fall-midterm/7/solution.md tags: - comp2804 - comp2804-midterm - - comp2804-newton's-binomial-theorem + - comp2804-the-pigeonhole-principle --- diff --git a/src/content/questions/comp2804/2015-fall-midterm/8/index.md b/src/content/questions/comp2804/2015-fall-midterm/8/index.md index 1eaad362..cdd81da6 100644 --- a/src/content/questions/comp2804/2015-fall-midterm/8/index.md +++ b/src/content/questions/comp2804/2015-fall-midterm/8/index.md @@ -8,5 +8,5 @@ solution: comp2804/2015-fall-midterm/8/solution.md tags: - comp2804 - comp2804-midterm - - comp2804-counting-solutions-of-linear-equations + - comp2804-newton's-binomial-theorem --- diff --git a/src/content/questions/comp2804/2015-fall-midterm/8/solution.md b/src/content/questions/comp2804/2015-fall-midterm/8/solution.md index 6066b823..178edb9b 100644 --- a/src/content/questions/comp2804/2015-fall-midterm/8/solution.md +++ b/src/content/questions/comp2804/2015-fall-midterm/8/solution.md @@ -1,4 +1,4 @@ -$=\sum^{88}_{k=0} \binom{88}{k}{(3x)}^k {(-17y)}^{88-k}$ +$=\sum^{88}_{k=0} \binom{88}{k}{(3x)}^{88-k} {(-17y)}^{k}$ $=\binom{88}{7}{(3x)}^{81} {(-17y)}^{7}$ diff --git a/src/content/questions/comp2804/2015-winter-final/4/solution.md b/src/content/questions/comp2804/2015-winter-final/4/solution.md index dd57f448..34455183 100644 --- a/src/content/questions/comp2804/2015-winter-final/4/solution.md +++ b/src/content/questions/comp2804/2015-winter-final/4/solution.md @@ -1,7 +1,13 @@ $ {(2x-7y)}^{15} $ -$= \sum\_{k=4}^{15} \binom{15}{k} {(2x)}^{k} {(-7y)}^{15-k} $ +$ = \sum_{k=0}^{15} \binom{15}{k} {(2x)}^{n-k} {(-7y)}^{k} $ -$ = \binom{15}{4} 2^{4} {(-7)}^{11} x^4 y^{11}$ +We only consider $k=11$, as it results in $y^{11}$. -$ = - \binom{15}{4} 2^{4} {(7)}^{11} $ +$ = \binom{15}{11} \cdot {(2x)}^{15-11} \cdot {(-7y)}^{11} $ + +$ = \binom{15}{11} \cdot 2^{4} \cdot {(-7)}^{11} \cdot x^4 \cdot y^{11} $ + +$ = - \binom{15}{4} \cdot 2^{4} \cdot 7^{11} \cdot x^4 \cdot y^{11} $ + +Thus, the coefficient of $ x^{4}y^{11} $ in the expansion of $ {(2x-7y)}^{15} $ is $ - \binom{15}{11} \cdot {2}^{4} \cdot {7}^{11} $ \ No newline at end of file diff --git a/src/content/questions/comp2804/2016-fall-midterm/9/solution.md b/src/content/questions/comp2804/2016-fall-midterm/9/solution.md index 4707b051..80b2dcce 100644 --- a/src/content/questions/comp2804/2016-fall-midterm/9/solution.md +++ b/src/content/questions/comp2804/2016-fall-midterm/9/solution.md @@ -1,7 +1,11 @@ -$ = \sum^{50}\_{k=0} \binom{50}{k} {(5x)}^{50-k} {(-7y)}^k $ +$ = \sum^{50}_{k=0} \binom{50}{k} {(5x)}^{n-k} {(-7y)}^k $ -$ = \sum^{50}\_{k=0} \binom{50}{26} {(5x)}^{50-26} {(-7y)}^k $ +We only consider $k=26$, as it results in $y^{26}$. -$ = \binom{50}{24} {(5)}^{24} x^{24} {(-7)}^{26} y^{26} $ +$ = \binom{50}{26} \cdot {(5x)}^{50-26} \cdot {(-7y)}^{26} $ -$ = \binom{50}{26} {(5)}^{24} {(-7)}^{26} x^{24} y^{26} $ +$ = \binom{50}{26} \cdot {(5)}^{24} \cdot x^{24} \cdot {(-7)}^{26} \cdot y^{26} $ + +$ = \binom{50}{26} \cdot 5^{24} \cdot 7^{26} \cdot x^{24} \cdot y^{26} $ + +Thus, the coefficient is $ \binom{50}{26} \cdot 5^{24} \cdot 7^{26} $ \ No newline at end of file diff --git a/src/content/questions/comp2804/2017-fall-midterm/9/solution.md b/src/content/questions/comp2804/2017-fall-midterm/9/solution.md index b18e9c50..cc40c4a9 100644 --- a/src/content/questions/comp2804/2017-fall-midterm/9/solution.md +++ b/src/content/questions/comp2804/2017-fall-midterm/9/solution.md @@ -1,11 +1,11 @@ -$ = \sum^{100}\_{k=0} \binom{100}{k} {(7x)}^{k} {(-13y)}^{100-k} $ +$ = \sum^{100}_{k=0} \binom{100}{k} {(7x)}^{n-k} {(-13y)}^{k} $ -$ = \sum^{100}\_{k=0} \binom{100}{20} {(7x)}^{20} {(-13y)}^{100-20} $ +We only consider $k=80$, as it results in $y^{80}$. -$ = \binom{100}{20} {(7)}^{20} x^{20} {(-13)}^{80} y^{80} $ +$ = \binom{100}{80} \cdot {(7x)}^{100-80} \cdot {(-13y)}^{80} $ -$ = \binom{100}{20} {(7)}^{20} {(-13)}^{80} x^{20} y^{80} $ +$ = \binom{100}{80} \cdot {(7)}^{20} \cdot x^{20} \cdot {(-13)}^{80} \cdot y^{80} $ -$ = \binom{100}{20} {(7)}^{20} {(13)}^{80} x^{20} y^{80} $ +$ = \binom{100}{80} \cdot 7^{20} \cdot 13^{80} \cdot x^{20} \cdot y^{80} $ -Thus, the coefficient is $ \binom{100}{20} {(7)}^{20} {(13)}^{80} $ +Thus, the coefficient is $ \binom{100}{80} \cdot 7^{20} \cdot 13^{80} $ diff --git a/src/content/questions/comp2804/2018-fall-final/6/solution.md b/src/content/questions/comp2804/2018-fall-final/6/solution.md index 9e01cbe4..cbf2a2b5 100644 --- a/src/content/questions/comp2804/2018-fall-final/6/solution.md +++ b/src/content/questions/comp2804/2018-fall-final/6/solution.md @@ -1,7 +1,11 @@ -$ = \sum\_{k = 35}^{55} \binom{55}{k} {(5x)}^{k} {(-3y)}^{n-k} $ +$ = \sum_{k=0}^{55} \binom{55}{k} {(5x)}^{n-k} {(-3y)}^{k} $ -$ = \binom{55}{20} {(5x)}^{20} {(-3y)}^{35} $ +We only consider $k=35$, as it results in $y^{35}$. -$ = - \binom{55}{20} 5^{20} 3^{35} x^{20} y^{35}$ +$ = \binom{55}{35} \cdot {(5x)}^{55-35} \cdot {(-3y)}^{35} $ -The coefficient is $ - \binom{55}{20} 5^{20} 3^{35} $ +$ = \binom{55}{35} \cdot 5^{20} \cdot {(-3)}^{35} \cdot x^{20} \cdot y^{35}$ + +$ = - \binom{55}{35} \cdot 5^{20} \cdot 3^{35} \cdot x^{20} \cdot y^{35}$ + +The coefficient is $ - \binom{55}{35} \cdot 5^{20} \cdot 3^{35} $ diff --git a/src/content/questions/comp2804/2022-winter-final/6/solution.md b/src/content/questions/comp2804/2022-winter-final/6/solution.md index ebe67177..713b22cd 100644 --- a/src/content/questions/comp2804/2022-winter-final/6/solution.md +++ b/src/content/questions/comp2804/2022-winter-final/6/solution.md @@ -1,13 +1,15 @@ $ (2x - 3y)^{30} $ -$ = \sum\_{k=10}^{30} \binom{30}{k} \cdot (2x)^{k} \cdot (-3y)^{30-k} $ +$ = \sum_{k=0}^{30} \binom{30}{k} \cdot (2x)^{n-k} \cdot (-3y)^{k} $ -$ = \binom{30}{10} \cdot (2x)^{10} \cdot (-3y)^{30-10} $ +We only consider $k=20$, as it results in $y^{20}$. -$ = \binom{30}{10} \cdot (2x)^{10} \cdot (-3y)^{20} $ +$ = \binom{30}{20} \cdot (2x)^{30-20} \cdot (-3y)^{20} $ -$ = \binom{30}{10} \cdot 2^{10} \cdot (-3)^{20} \cdot x^{10} \cdot y^{20} $ +$ = \binom{30}{20} \cdot (2x)^{10} \cdot (-3y)^{20} $ -$ = \binom{30}{10} \cdot 2^{10} \cdot (3)^{20} \cdot x^{10} \cdot y^{20} $ +$ = \binom{30}{20} \cdot 2^{10} \cdot (-3)^{20} \cdot x^{10} \cdot y^{20} $ -From this equation, we can see that the coefficient (aka the real numbers) are: $\binom{30}{10} \cdot 2^{10} \cdot (3)^{20}$ +$ = \binom{30}{20} \cdot 2^{10} \cdot (3)^{20} \cdot x^{10} \cdot y^{20} $ + +From this equation, we can see that the coefficient (aka the real numbers) are: $\binom{30}{20} \cdot 2^{10} \cdot 3^{20}$