If an=43−(43)2+(43)3+...+(−1)n−1(43)n and bn=1−an, then find the least natural number n0 such that bn>an∀n≥n0.

an =43−(43)3+(43)3+⋯+(−1)n−1(43)n=1−(43)43(1−(43)n)=73(1−(−43)n)Now, bn=1−an and bn>an for n≥n0.∴ 1−an>an ⇒2an<1⇒76[1−(−43)n]<1⇒−(−43)n<61⇒(−3)n+1<22n−1For n to be even, inequality always holds. For n to be odd, it holds for n≥7. Therefore, the least natural number for which it holds is 6.

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If a_n = {3}{4} - ({3}{4})^2 + ({3}{4})^3 + ... + (-1)^{n - 1}

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If $a_{n} = \frac{3}{4} - (\frac{3}{4})^{2} + (\frac{3}{4})^{3} + ... + (- 1)^{n - 1} (\frac{3}{4})^{n}$ and $b_{n} = 1 - a_{n}$ , then find the least natural number $n_{0}$ such that $b_{n} > a_{n} \forall n \geq n_{0}$ .

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a_{n} = \frac{3}{4} - (\frac{3}{4})^{2} + (\frac{3}{4})^{3} + ... + (- 1)^{n - 1} (\frac{3}{4})^{n}

and

b_{n} = 1 - a_{n}

, then find the least natural number

n_{0}

such that

b_{n} > a_{n} \forall n \geq n_{0}

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Text solutionVerified

a_{n} = \frac{3}{4} - (\frac{3}{4})^{3} + (\frac{3}{4})^{3} + \dots + (- 1)^{n - 1} (\frac{3}{4})^{n}

= \frac{\frac{3}{4} ( 1 - ( \frac{3}{4} ) ^{n} )}{1 - ( \frac{3}{4} )}

= \frac{3}{7} (1 - (- \frac{3}{4})^{n})

Now,

b_{n} = 1 - a_{n}

and

b_{n} > a_{n}

for

n \geq n_{0}

∴

1 - a_{n} > a_{n} \Rightarrow 2 a_{n} < 1

\Rightarrow \frac{6}{7} [1 - (- \frac{3}{4})^{n}] < 1

\Rightarrow - (- \frac{3}{4})^{n} < \frac{1}{6}

\Rightarrow (- 3)^{n + 1} < 2^{2 n - 1}

For n to be even, inequality always holds. For n to be odd, it holds for

n \geq 7

. Therefore, the least natural number for which it holds is 6.

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Question Text	If $a_{n} = \frac{3}{4} - (\frac{3}{4})^{2} + (\frac{3}{4})^{3} + ... + (- 1)^{n - 1} (\frac{3}{4})^{n}$ and $b_{n} = 1 - a_{n}$ , then find the least natural number $n_{0}$ such that $b_{n} > a_{n} \forall n \geq n_{0}$ .
Answer Type	Text solution:1
Upvotes	150

If an​=43​−(43​)2+(43​)3+...+(−1)n−1(43​)n and bn​=1−an​, then find the least natural number n0​ such that bn​>an​∀n≥n0​.

Text solutionVerified

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If $a_{n} = \frac{3}{4} - (\frac{3}{4})^{2} + (\frac{3}{4})^{3} + ... + (- 1)^{n - 1} (\frac{3}{4})^{n}$ and $b_{n} = 1 - a_{n}$ , then find the least natural number $n_{0}$ such that $b_{n} > a_{n} \forall n \geq n_{0}$ .