Question

Prove by mathematical induction.
4ⁿ ≥ (1 + 3n)
for n ≥ 1

Answer 1

$\large\underline{\sf{Solution-}}$

Let assume that

$\rm \: P(n): \: {4}^{n} \geqslant 1 + 3n$

Step :- 1 For n = 1

$\rm \: P(1): \: {4}^{1} \geqslant 1 + 3 \times 1$

$\rm\implies \:P(1):4 = 4$

$\rm\implies \:P(1) \: is \: true \: for \: n = 1$

Step :- 2 Let assume that P(n) is true for n = k, where k is some natural number.

$\rm \: P(k): \: {4}^{k} \geqslant 1 + 3k - - - (1)$

Step :- 3 Now, we have to prove that P(n) is true for n = k + 1

$\color{blue}\rm \: P(k + 1): \: {4}^{k + 1} \geqslant 1 + 3(k + 1)$

Now, from equation (1), we have

$\rm \: \: {4}^{k} \geqslant 1 + 3k$

On multiply by 4, on both sides we get

$\rm \: \: 4 \times {4}^{k} \geqslant 4(1 + 3k)$

$\rm \: \: {4}^{k + 1} \geqslant 4 + 12k$

can be rewritten as

$\rm \: \: {4}^{k + 1} \geqslant 1 + 3 + 3k + 9k > 1 + 3 + 3k$

$\rm \: \: {4}^{k + 1} \geqslant 1 + 3(1 + k)$

$\rm\implies \:\rm \: \: {4}^{k + 1} \geqslant 1 + 3(k + 1)$

$\rm\implies \:P(n) \: is \: true \: for \: n = k + 1$

Hence, By the Principle of Mathematical Induction,

$\color{green}\rm\implies \:\rm \: \: {4}^{n} \geqslant 1 + 3n \: \: \forall \: n \geqslant 1$