Question

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

Answer 1

Step-by-step explanation:

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.

✅Hope it helps you ❗