If a - b - 1- then an-bn-na-b for every - ive integer n - 2-

A/b gt;1 ∴ 1+a/b+a^2/b^2+ …… ( a/b )^n 1 gt;n as each is gt;1 Now sum a G.P. in L.H.S etc a^n b^n/b^n 1(a b) gt;n or a^n b^n gt; n(a b)b^n 1 gt;n(a ...

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Standard XII

Mathematics

Inequality

If a > b > 1,...

Question

If a > b > 1, then $a^{n} - b^{n} > n (a - b)$ for every + ive integer $n \geq 2$ .

True

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False

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Solution

The correct option is A True
$\frac{a}{b} > 1$
$∴ 1 + \frac{a}{b} + \frac{a^{2}}{b^{2}} + \dots \dots {(\frac{a}{b})}^{n - 1} > n$ as each is >1
Now sum a G.P. in L.H.S etc
$\frac{a^{n} - b^{n}}{b^{n - 1} (a - b)} > n$
or $a^{n} - b^{n} > n (a - b) b^{n - 1}$
$> n (a - b)$
as $b > 1 \Rightarrow b^{n - 1} > 1$

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If a > b > 1, then an−bn>n(a−b) for every + ive integer n≥2.

The correct option is A Trueab>1 ∴1+ab+a2b2+……(ab)n−1>n as each is >1 Now sum a G.P. in L.H.S etc an−bnbn−1(a−b)>n or an−bn>n(a−b)bn−1 >n(a−b) as b>1⇒bn−1>1

If a > b > 1, then $a^{n} - b^{n} > n (a - b)$ for every + ive integer $n \geq 2$ .