Prove-that-4n-lt-2-n-for-all-n-5-

Question Number 176102 by MathsFan last updated on 12/Sep/22

Prove that

4 n < 2^{n} for all n ⩾ 5

Answered by a.lgnaoui last updated on 12/Sep/22

n = 5 2^{5} = 32 4 \times 5 = 20 < 32

n = 6 2^{6} = 64 4 \times 6 = 24 < 64

n = 7 2^{7} = 128 4 \times 7 = 28 < 128

\dots \dots \dots \dots \dots \dots \dots \dots \dots

n + 1 2^{n + 1} 4 (n + 1) = 4 n + 4 < 2^{n} + 4

n [4 (n + 1)] < n (2^{n} + 4) < n (2 \times 2^{n}) < n 2^{n + 1}

d o n c 4 (n + 1) < 2^{n + 1}

n e s t v r a i p o u r t o u t n ⩾ 5 (n \in N)

Commented by MathsFan last updated on 13/Sep/22

Merci Monsieur

Answered by mahdipoor last updated on 12/Sep/22

i f 4 a < 2^{a} \Rightarrow

4 a + 4 < 2^{a} + 4 = 2^{a} \times (1 + \frac{4}{2^{a}}) < 2^{a} \times 2 (a > 5)

\Rightarrow 4 (a + 1) < 2^{a + 1}

a n d 4 \times 5 < 2^{5}

\Rightarrow\Rightarrow 4 n < 2^{n} f o r \forall n ⩾ 5

Commented by MathsFan last updated on 13/Sep/22

thank you sir