Question

Let N = ²⁰²¹C1 + 2.²⁰²¹ C2 + 3. ²⁰²¹ C3 + ... + 2021.
²⁰²¹C2021. Number of prime factors of N is equal to
opt.a) 8083
opt.a) 8
opt.a) 3
opt.a) 2​

Answer 1

Answer:

Here, abx

2

+(b

2

−ac)x−bc=0

x=

2A

−B±

(B

2

−4AC)

⇒x=

2ab

−(b

2

−ac)±

(b

2

−ac)

2

−4ab(−bc)

⇒x=

2ab

−(b

2

−ac)±

(b

2

−ac)

2

+4ab

2

c

⇒x=

2ab

−(b

2

−ac)±

b

4

−2ab

2

c+a

2

c

2

+4ab

2

c

⇒x=

2ab

−(b

2

−ac)±

(b

2

+ac)

2

⇒x=

2ab

−(b

2

−ac)+(b

2

+ac)

,x=

2ab

−(b

2

−ac)−(b

2

+ac)

⇒x=

2ab

2ac

,x=

2ab

−2b

2

⇒x=

b

c

,x=−

a

b

Answer 2

Answer:

option C is correct answer