If a gt- b gt- 1- then a-n-b-n gt-n-a-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 ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Mathematics

Inequality

If a gt; b...

Question

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

True

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

False

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

$\frac{a}{b} > 1$
$∴ 1 + \frac{a}{b} + \frac{a^{2}}{b^{2}} + \dots \dots {(\frac{a}{b})}^{n - 1} > n$ 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 $b > 1 \Rightarrow b^{n - 1} > 1$

Suggest Corrections

Similar questions

Q. If A = diag (a, b, c), show that Aⁿ = diag (aⁿ, bⁿ, cⁿ) for all positive integer n.

Q. For some positive integer n, every positive odd integer is of the form

(a) n
(b) n + 1
(c) 2n
(d) 2n + 1

If a and b are distinct integers, prove that a – b is a factor of aⁿ – bⁿ, whenever n is a positive integer.

[Hint: write aⁿ = (a – b + b)ⁿ and expand]

Q. If a > b > 1, then

a^{n} - b^{n} > n (a - b)

for every + ive integer

n \geq 2

Q. If a > b and n is a + ive integer, then

a^{n} - b^{n} > n (a b)^{\frac{(n - 1)}{2}} (a - b)

Join BYJU'S Learning Program

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

ab>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 $n \geq 2$ .

$\frac{a}{b} > 1$
$∴ 1 + \frac{a}{b} + \frac{a^{2}}{b^{2}} + \dots \dots {(\frac{a}{b})}^{n - 1} > n$ 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 $b > 1 \Rightarrow b^{n - 1} > 1$