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Let $a$ and $b$ be positive integers. Which of the following statements is true for all positive integers $n$ ?
Option: 1
${{(a+b)}^{n}}>{{a}^{n}}+{{b}^{n}}$

Option: 2
${{a}^{n}}+{{b}^{n}}>{{(a+b)}^{n}}$

Option: 3
${{(a+b)}^{n}}={{a}^{n}}+{{b}^{n}}$

Option: 4
${{(a+b)}^{n}}<{{a}^{n}}+{{b}^{n}}$

Answers (1)

best_answer

We will use the principle of mathematical induction to solve this problem.

The base case is when $n=1$ , then ${{(a+b)}^{1}}>{{a}^{1}}+{{b}^{1}}$ which is the correct value.

Therefore, ${{(a+b)}^{n}}>{{a}^{n}}+{{b}^{n}}$

is true for n = 1.Now, assume that the statement is true for some positive integer k.

${{(a+b)}^{k}}>{{a}^{k}}+{{b}^{k}$

We need to show that

$\left ( a+b^{} \right )^{\left ( k+1 \right )}> a^{\left ( k+1 \right )}+b^{\left ( k+1 \right )}$

Expanding the left-hand side using the binomial theorem,

${{(a+b)}^{(k+1)}}={{(a+b)}^{k}}\times (a+b) \\$

${{(a+b)}^{(k+1)}}=({{a}^{k}}+k{{a}^{(k-1)}}b+...)\times (a+b)$

${{(a+b)}^{(k+1)}}={{a}^{(k+1)}}+(k+1){{a}^{k}}b+...$

Since $a$ and $b$ are positive and we know that,

$k{{a}^{(k-1)}}b<{{a}^{k}}$ and $k{{b}^{k}}<{{b}^{(k+1)}}$

Therefore, we have:

${{(a+b)}^{(k+1)}}>{{a}^{(k+1)}}+(k+1){{a}^{k}}b+k{{b}^{(k+1)}} \\$

${{(a+b)}^{(k+1)}}>{{a}^{(k+1)}}+{{b}^{(k+1)}} \\$

Therefore, the statement is true for all positive integers $n$ .

Hence, ${{(a+b)}^{n}}>{{a}^{n}}+{{b}^{n}}$

is true for all positive integers $n$ .

Posted by

Ritika Kankaria

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