2015 Countdown Problem #4: Find the Largest - Part I

Algebra Level 4

Which of the following is the largest?

A. 2 11 2 11 2 11 2 11 . . . − \sqrt{2^{11}\sqrt{2^{11}\sqrt{2^{11}\sqrt{2^{11}\sqrt{...}}}}}- 33 33 33 33 . . . \sqrt{33\sqrt{33\sqrt{33\sqrt{33\sqrt{...}}}}}

B. 1 2 + 1 1 + 201 4 − 2014 + 1 1 + 201 3 − 2013 + 1 1 + 201 2 − 2012 + . . . \frac{1}{2}+\frac{1}{1+2014^{-2014}}+\frac{1}{1+2013^{-2013}}+\frac{1}{1+2012^{-2012}}+... + 1 1 + 201 2 2012 + 1 1 + 201 3 2013 + 1 1 + 201 4 2014 +\frac{1}{1+2012^{2012}}+\frac{1}{1+2013^{2013}}+\frac{1}{1+2014^{2014}}

C. 1008 × ( 1 − 1 2 2 ) − 1 × ( 1 − 1 3 2 ) − 1 × ( 1 − 1 4 2 ) − 1 . . . 1008\times(1-\frac{1}{2^{2}})^{-1}\times(1-\frac{1}{3^{2}})^{-1}\times(1-\frac{1}{4^{2}})^{-1}... ( 1 − 1 201 3 2 ) − 1 × ( 1 − 1 201 4 2 ) − 1 × ( 1 − 1 201 5 2 ) − 1 (1-\frac{1}{2013^{2}})^{-1}\times(1-\frac{1}{2014^{2}})^{-1}\times(1-\frac{1}{2015^{2}})^{-1}

This problem is part of the set 2015 Countdown Problems .

B is the largest. They are all equal. A is the largest. C is the largest.

Wee Xian Bin by Wee Xian Bin , 25, Singapore

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1 solution

Wee Xian Bin
Dec 12, 2014

For (a):

y = k y → y 2 = k y y=\sqrt{ky}\rightarrow y^2=ky

y = k y=k (reject y=0)

Value = 2 11 − 33 = 2015 2^{11}-33 = 2015

For (b):

1 ( 1 + 1 k n ) + 1 ( 1 + k n ) = k n ( k n + 1 ) + 1 ( 1 + k n ) = 1 \frac{1}{(1+\frac{1}{k^n})}+\frac{1}{(1+k^n)}=\frac{k^n}{(k^n+1)}+\frac{1}{(1+k^n)}=1

hence the sum is independent of both k and n.

Note the lone term 1 1 + ( 2012 ) 0 = 1 2 \frac{1}{1+(2012)^0}=\frac{1}{2} in the series is not part of any pair.

Sum = 1 2 + 2014 × 1 + 1 2 = 2015 \frac{1}{2}+2014×1+\frac{1}{2} = 2015

For (c):

( 1 − 1 k 2 ) − 1 = ( ( k + 1 ) ( k − 1 ) k 2 ) − 1 (1-\frac{1}{k^2})^{-1}=(\frac{(k+1)(k-1)}{k^2})^{-1} = k 2 ( k + 1 ) ( k − 1 ) = k ( k + 1 ) × k ( k − 1 ) =\frac{k^2}{(k+1)(k-1)} =\frac{k}{(k+1)}×\frac{k}{(k-1)}

1008 ( 1 − 1 2 2 ) − 1 ( 1 − 1 3 2 ) − 1 … ( 1 − 1 2014 2 ) − 1 ( 1 − 1 2015 2 ) − 1 1008(1-\frac{1}{{2}^2} )^{-1} (1-\frac{1}{{3}^2})^{-1}…(1-\frac{1}{{2014}^2})^{-1} (1-\frac{1}{{2015}^2})^{-1}

= 1008 ( 2 3 × 2 1 ) ( 3 4 × 3 2 ) … ( 2014 2015 × 2014 2013 ) ( 2015 2016 × 2015 2014 ) =1008(\frac{2}{3}×\frac{2}{1})(\frac{3}{4}×\frac{3}{2})…(\frac{2014}{2015}×\frac{2014}{2013})(\frac{2015}{2016}×\frac{2015}{2014})

= 1008 ( 2 3 × 3 4 × … × 2014 2015 × 2015 2016 ) ( 2 1 × 3 2 × … × 2014 2013 × 2015 2014 ) =1008(\frac{2}{3}×\frac{3}{4}×…×\frac{2014}{2015}×\frac{2015}{2016})(\frac{2}{1}×\frac{3}{2}×…×\frac{2014}{2013}×\frac{2015}{2014})

= 1008 ( 2 2016 ) ( 2015 ) =1008(\frac{2}{2016})(2015)

= 2015 =2015

2 Helpful 0 Interesting 0 Brilliant 0 Confused

Same way genius!! great solution

Kaustubh Miglani - 5 years, 3 months ago

Note: for B, I was unsure if by your series that if 1 1 + 0 0 \frac{1}{1+0^{0}} was going to be part of your summation, it would be great to clarify (although the option can be solved just from figuring out A=C)

Jared Low - 6 years, 5 months ago

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