fix: lint in main

src/content/questions/comp2804/2014-fall-midterm/7/solution.md

@@ -1,6 +1,6 @@
 ${(5x-36)}^{100}$
 
-$ = \sum^{100}_{k=0} \binom{100}{k} {(5x)}^{n-k} {(-3y)}^{k} $
+$ = \sum^{100}\_{k=0} \binom{100}{k} {(5x)}^{n-k} {(-3y)}^{k} $
 
 We only consider $k=80$, as it results in $y^{80}$.
 

src/content/questions/comp2804/2015-fall-final/4/solution.md

@@ -1,4 +1,4 @@
-$ = \sum^{20}_{k=0} \binom{20}{k} {(-3x)}^{n-k} {(5y)}^{k} $
+$ = \sum^{20}\_{k=0} \binom{20}{k} {(-3x)}^{n-k} {(5y)}^{k} $
 
 We only consider $k=5$, as it results in $y^{5}$.
 

src/content/questions/comp2804/2015-winter-final/4/solution.md

@@ -1,6 +1,6 @@
 $ {(2x-7y)}^{15} $
 
-$ = \sum_{k=0}^{15} \binom{15}{k} {(2x)}^{n-k} {(-7y)}^{k} $
+$ = \sum\_{k=0}^{15} \binom{15}{k} {(2x)}^{n-k} {(-7y)}^{k} $
 
 We only consider $k=11$, as it results in $y^{11}$.
 
@@ -10,4 +10,4 @@ $ = \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} $
+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} $

src/content/questions/comp2804/2016-fall-midterm/9/solution.md

@@ -1,4 +1,4 @@
-$ = \sum^{50}_{k=0} \binom{50}{k} {(5x)}^{n-k} {(-7y)}^k $
+$ = \sum^{50}\_{k=0} \binom{50}{k} {(5x)}^{n-k} {(-7y)}^k $
 
 We only consider $k=26$, as it results in $y^{26}$.
 
@@ -8,4 +8,4 @@ $ = \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} $
+Thus, the coefficient is $ \binom{50}{26} \cdot 5^{24} \cdot 7^{26} $

src/content/questions/comp2804/2017-fall-midterm/9/solution.md

@@ -1,4 +1,4 @@
-$ = \sum^{100}_{k=0} \binom{100}{k} {(7x)}^{n-k} {(-13y)}^{k} $
+$ = \sum^{100}\_{k=0} \binom{100}{k} {(7x)}^{n-k} {(-13y)}^{k} $
 
 We only consider $k=80$, as it results in $y^{80}$.
 

src/content/questions/comp2804/2018-fall-final/6/solution.md

@@ -1,4 +1,4 @@
-$ = \sum_{k=0}^{55} \binom{55}{k} {(5x)}^{n-k} {(-3y)}^{k} $
+$ = \sum\_{k=0}^{55} \binom{55}{k} {(5x)}^{n-k} {(-3y)}^{k} $
 
 We only consider $k=35$, as it results in $y^{35}$.
 

src/content/questions/comp2804/2022-winter-final/6/solution.md

@@ -1,6 +1,6 @@
 $ (2x - 3y)^{30} $
 
-$ = \sum_{k=0}^{30} \binom{30}{k} \cdot (2x)^{n-k} \cdot (-3y)^{k} $
+$ = \sum\_{k=0}^{30} \binom{30}{k} \cdot (2x)^{n-k} \cdot (-3y)^{k} $
 
 We only consider $k=20$, as it results in $y^{20}$.