Least odd natural number

Algebra Level 5

{ A n = ( 3 4 ) − ( 3 4 ) 2 + ( 3 4 ) 3 + ⋯ + ( − 1 ) n − 1 ( 3 4 ) n B n = 1 − A n \begin{cases} A_n=\left(\frac{3}{4}\right)-\left(\frac{3}{4}\right)^2+\left(\frac{3}{4}\right)^3+\cdots+(-1)^{n-1}\left(\frac{3}{4}\right)^n \\ B_n=1-A_n \end{cases}

Find the least odd positive integer, n 0 n_0 such that B n > A n B_n>A_n for all n ≥ n 0 n \geq n_0 .

You can try my other Sequences And Series problems by clicking here : Part II and here : Part I.


The answer is 7.

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Sandeep Bhardwaj by Sandeep Bhardwaj , 28, India

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1 solution

Chew-Seong Cheong
Dec 18, 2014

A n = ( 3 4 ) − ( 3 4 ) 2 + ( 3 4 ) 3 − . . . + ( − 1 ) n − 1 ( 3 4 ) n A_n = \left( \frac {3}{4} \right) - \left( \frac {3}{4} \right)^2 + \left( \frac {3}{4} \right)^3 - ... + (-1)^{n-1} \left( \frac {3}{4} \right)^n

For an odd n n ,

A n = ( 3 4 ) + ( 3 4 ) 2 + ( 3 4 ) 3 + . . . + ( 3 4 ) n A_n = \left( \frac {3}{4} \right) + \left( \frac {3}{4} \right)^2 + \left( \frac {3}{4} \right)^3 + ... + \left( \frac {3}{4} \right)^n − 2 [ ( 3 4 ) 2 + ( 3 4 ) 4 + ( 3 4 ) 6 + . . . + ( − 1 ) n − 1 ( 3 4 ) n − 1 ] \quad \quad \quad \quad - 2 \left[ \left( \frac {3}{4} \right)^2 + \left( \frac {3}{4} \right)^4 + \left( \frac {3}{4} \right)^6 + ... + (-1)^{n-1} \left( \frac {3}{4} \right)^{n-1} \right]

= 3 4 [ 1 + ( 3 4 ) 2 + ( 3 4 ) 4 + . . . + ( − 1 ) n − 1 ( 3 4 ) n − 1 ] \quad \space \space = \frac {3}{4} \left[ 1 + \left( \frac {3}{4} \right)^2 + \left( \frac {3}{4} \right)^4 + ... + (-1)^{n-1} \left( \frac {3}{4} \right)^{n-1} \right] − 2 ( 3 4 ) 2 [ ( 1 + 3 4 ) 2 + ( 3 4 ) 4 + . . . + ( − 1 ) n − 1 ( 3 4 ) n − 2 ] \quad \quad \quad \quad - 2 \left( \frac {3}{4} \right)^2 \left[ \left( 1 + \frac {3}{4} \right)^2 + \left( \frac {3}{4} \right)^4 + ... + (-1)^{n-1} \left( \frac {3}{4} \right)^{n-2} \right]

= 3 4 [ 1 − ( 3 4 ) n 1 − ( 3 4 ) ] − 2 ( 9 16 ) [ 1 − ( 3 4 ) n − 1 1 − ( 9 16 ) ] \quad \space \space = \frac {3}{4} \left[ \dfrac {1 - \left( \frac {3}{4} \right)^n } {1 - \left( \frac {3}{4} \right)} \right] - 2 \left( \frac {9}{16} \right) \left[ \dfrac {1- \left( \frac {3}{4} \right)^{n-1}} {1- \left( \frac {9}{16} \right) } \right]

= 3 [ 1 − ( 3 4 ) n ] − 18 7 [ 1 − ( 3 4 ) n − 1 ] \quad \space \space = 3 \left[ 1 - \left( \frac {3}{4} \right)^n \right] - \frac {18}{7} \left[ 1 - \left(\frac {3}{4} \right)^{n-1} \right]

= 3 − 3 ( 3 4 ) n − 18 7 + 18 7 ( 3 4 ) n − 1 \quad \space \space = 3 - 3 \left( \frac {3}{4} \right)^n - \frac {18}{7} + \frac {18}{7} \left(\frac {3}{4} \right)^{n-1}

= 3 7 + 9 28 ( 3 4 ) n − 1 \quad \space \space = \frac {3}{7} + \frac {9}{28} \left(\frac {3}{4} \right)^{n-1}

For B n > A n ⇒ 1 − A n > A n ⇒ A n < 1 2 B_n > A_n\quad \Rightarrow 1 - A_n > A_n \quad \Rightarrow A_n < \frac {1} {2}

⇒ 3 7 + 9 28 ( 3 4 ) n − 1 < 1 2 ⇒ ( 3 4 ) n − 1 < 2 9 \Rightarrow \frac {3}{7} + \frac {9}{28} \left(\frac {3}{4} \right)^{n-1} < \frac {1}{2} \quad \Rightarrow \left(\frac {3}{4} \right)^{n-1} < \frac {2}{9}

⇒ ( n − 1 ) log ⁡ ( 3 4 ) < log ⁡ ( 2 9 ) \Rightarrow (n-1) \log {\left(\frac {3}{4} \right)} < \log {\left(\frac {2}{9} \right) }

⇒ − 0.124938737 ( n − 1 ) < − 0.653212514 \Rightarrow -0.124938737 (n-1) < -0.653212514

⇒ n − 1 > 5.228262519 ⇒ n > 7 \Rightarrow n - 1 > 5.228262519 \quad \Rightarrow n > \boxed {7}

14 Helpful 0 Interesting 0 Brilliant 0 Confused

Awesome solution. . as usual.

Sandeep Bhardwaj - 6 years, 5 months ago

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Sir I'am gotta confused with the answer I think n o { n }_{ o } . should be 5

Can you explain me it please ? Thanks Sir's @Sandeep Bhardwaj @Chew-Seong Cheong

Karan Shekhawat - 6 years, 5 months ago

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Sorry, I think your confusion comes from:

⇒ ( n − 1 ) log ⁡ ( 3 4 ) \Rightarrow (n-1) \log {\left(\frac {3}{4} \right)} " > > " log ⁡ ( 2 9 ) \log {\left(\frac {2}{9} \right) }

That was a typo. It should be " < < ". I have edited it.

⇒ ( n − 1 ) log ⁡ ( 3 4 ) < log ⁡ ( 2 9 ) \Rightarrow (n-1) \log {\left(\frac {3}{4} \right)} < \log {\left(\frac {2}{9} \right) }

⇒ − 0.124938737 ( n − 1 ) < − 0.653212514 \Rightarrow -0.124938737 (n-1) < -0.653212514

Here when you multiply both sides with − 1 -1 , the inequality sign changes from " < < " to " > > ".

⇒ + 0.124938737 ( n − 1 ) > + 0.653212514 \Rightarrow +0.124938737 (n-1) > +0.653212514

⇒ n − 1 > 5.228262519 ⇒ n > 7 \Rightarrow n - 1 > 5.228262519 \quad \Rightarrow n > \boxed {7}

Chew-Seong Cheong - 6 years, 5 months ago

I COULD NOT UNDERSTAND THE FIRST STEP. can you explain? please

ramesh perumal - 6 years, 5 months ago

Actually, $B 6>A 6$ - you are welcome to check that...

Dennis Gulko - 4 years, 4 months ago

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