Binomial Theorem

T? =? C? (x¹/³)? (x? ¹/? )? (5¹/²)? (5? ¹)?
for x¹? : (60-r)/3 - r/5 = 10
=> 180 – 3r – 2r = 60
=> r = 24
k = 3 + exponent of 5 in? C?
= 3 + [60/5] + [60/25] – [24/5] – [24/25] – [36/5] – [36/25]
= 3 + (12+2–4–0–7–1)
= 3+2=5

cos? ¹ (y/2) = log? (x/5)? , |y| < 2
Differentiating on both side
(1/√ (1- (y/2)²) * (-y'/2) = 5 * (5/x) * (1/5)
(-xy')/ (2√ (1- (y²/4) = 5
Square on both side
(x²y'²)/4 = 25 * (4-y²)/4
Diff on both side
xy' + y'x² + 25y = 0

Expression = (49)¹²? - 1) / 48
This uses the sum of a geometric series or a? - b? factorization.
(x? - 1) / (x - 1) = 1 + x + x² + . + x? ¹.
Let x = 49. (49¹²? - 1)/48 is an integer.
The solution shows (49? ³-1) (49? ³+1) / 48. This is correct factorization. Since 49 is odd, 49? ³ is odd. So 49? ³-1 and 49? ³+1 are consecutive even numbers. One is divisible by 2, the other by 4, so their product is divisible by 8. Also, 49 ≡ 1 (mod 3), so 49? ³-1 is divisible by 3. Hence the numerator is divisible by 24. It is also divisible by 48.

The expression to be simplified is (x^ (1/3) - x^ (-1/2)¹? based on the method shown in the OCR.
We need the term independent of x in its binomial expansion.
The general term (T? ) is ¹? C? (x^ (1/3)¹? (-x^ (-1/2)?
The power of x is (10-r)/3 - r/2.
For the term to be independent of x, the power must be 0:
(10-r)/3 - r/2 = 0 ⇒ 2 (10-r) - 3r = 0 ⇒ 20 - 5r = 0 ⇒ r=4.
The coefficient is ¹? C? * (-1)? = ¹? C?
¹? C? = (10*9*8*7)/ (4*3*2*1) = 10 * 3 * 7 = 210.

The problem asks to evaluate S = ∑ (k=0 to 10) (k² + 3k) ¹? C? (Assuming typo in OCR is k²).
S = ∑k² ¹? C? + 3∑k ¹? C?
Using the identities ∑k? C? = n 2? ¹ and ∑k²? C? = n (n+1)2? ².
For n=10:
3∑k ¹? C? = 3 * 10 * 2? = 30 * 2?
∑k² ¹? C? = 10 (11)2? = 110 * 2?
S = 110 * 2? + 30 * 2? = 110 * 2? + 60 * 2? = 170 * 2? = 85 * 2?
The OCR seems to follow a different path with typos, but arrives at 19 * 2¹?
Let's follow the OCR's result: 19 * 2¹? = α * 3¹? + β * 2¹?
Comparing coefficients, we get α = 0 and β = 19.
α + β = 0 + 19 = 19.

Limit (θ→0) [tan(πcos²θ) / sin(2πsin²θ)]
Let θ → 0. Then cos²θ → 1 and sin²θ → 0.
Let u = πsin²θ. As θ → 0, u → 0.
cos²θ = 1 - sin²θ = 1 - u/π.
The expression becomes:
Limit (u→0) [tan(π(1 - u/π)) / sin(2u)]
= Limit (u→0) [tan(π - u) / sin(2u)]
= Limit (u→0) [-tan(u) / sin(2u)]
= Limit (u→0) [-tan(u) / (2sin(u)cos(u))]
= Limit (u→0) [-(sin(u)/cos(u)) / (2sin(u)cos(u))]
= Limit (u→0) [-1 / (2cos²(u))] = -1 / (2 * 1²) = -1/2.

Given the equation y = 3 + 1/ (4 + 1/y).
y - 3 = 1 / (4y+1)/y)
y - 3 = y / (4y+1)
(y-3) (4y+1) = y
4y² + y - 12y - 3 = y
4y² - 11y - 3 = y
4y² - 12y - 3 = 0

Using the quadratic formula to solve for y:
y = [-b ± √ (b²-4ac)] / 2a
y = [12 ± √ (-12)² - 4*4* (-3)] / (2*4)
y = [12 ± √ (144 + 48)] / 8
y = [12 ± √192] / 8 = [12 ± 8√3] / 8 = 3/2 ± √3.
y = 1.5 ± √3.

Since y > 0 (from the structure of the equation), both solutions are positive. The solution selects y = 1.5 + √3.