Question

Prove that 2n+1 > (n + 2) · sin(n) for all positive integers n.

Answer 1

Given :  2n+1 > (n + 2) · sin(n)  

To Find  : Prove above identity for all positive integers n.

Solution:

2n+1 > (n + 2) · sin(n)  

As Range of Sinx  is [-1, 1]

hence maximum value of sin (n) is  1  

=> 2n + 1 >  n + 2

=> n > 1

Hence its satisfied for all values of n

now lets check for n = 1

LHS = 2(1) + 1 = 3

RHS = (1 + 2) Sin(1)  = 3 Sin(1)

1  < π /2

Hence Sin 1  <  Sin π/2

=> Sin <  1

Hence RHS <  3(1)

=> LHS > RHS

Hence 2n+1 > (n + 2) · sin(n)    for n  = 1 also .

Hence 2n+1 > (n + 2) · sin(n) for all positive integers n.

Learn More:

Applying a suitable identity find the product of (x-y),(x+y),(x square+y ...

brainly.in/question/13813817

If 16a2 + 25b2- c2 = 40ab, then the family of lines ax + by + c = 0 is ...

brainly.in/question/14999769

Answer 2

Answer:

2n+1> (n+2) for n>1.

Since ∣sin(n)∣ ≤ 1, 2n+1> (n+2) sin(n) for n>1.

For n=1, sin(1)<1, & 3 > 3*sin(1) .

Therefore, 2n+1 > (n + 2) · sin(n) for all positive integers n.