Sum it up

Algebra Level 2

If the first three terms of a geometric progression are given to be 2 + 1 , 1 , 2 − 1 , \sqrt2+1,1,\sqrt2-1, find the sum to infinity of all of its terms.

If the answer is in the form of a + b c d \frac{a+b\sqrt c}d for positive integers a , b , c , a,b,c, and d d with c c square-free, find the minimum value of a + b + c + d a+b+c+d .


The answer is 11.

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Akshat Sharda by Akshat Sharda , 21, India

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3 solutions

Chew-Seong Cheong
Aug 15, 2015

The common ratio of the GP is r = 2 − 1 r = \sqrt{2}-1 . As ( 2 + 1 ) ( 2 − 1 ) = 2 − 1 = 1 (\sqrt{2}+1)( \sqrt{2}-1) = 2 - 1 = 1 , 1 ( 2 − 1 ) = ( 2 − 1 ) 1( \sqrt{2}-1) = (\sqrt{2}-1) ... Therefore,

( 2 + 1 ) + 1 + ( 2 − 1 ) + . . . = 2 + 1 1 − ( 2 − 1 ) = 2 + 1 2 − 2 = ( 2 + 1 ) ( 2 + 2 ) ( 2 − 2 ) ( 2 + 2 ) = 4 + 3 2 2 \begin{aligned} \left( \sqrt{2}+1 \right) + 1 + \left( \sqrt{2}-1 \right) + ... & = \frac{\sqrt{2}+1}{1-\left( \sqrt{2} - 1 \right)} \\ & = \frac{\sqrt{2}+1}{2- \sqrt{2}} \\ & = \frac{\left(\sqrt{2}+1\right) \left(2 + \sqrt{2} \right)}{\left(2- \sqrt{2}\right) \left(2 + \sqrt{2} \right)} \\ & = \frac{4+3\sqrt{2}}{2} \end{aligned}

⇒ a + b + c + d = 4 + 3 + 2 + 2 = 11 \Rightarrow a + b + c + d = 4 + 3 + 2 + 2 = \boxed{11}

26 Helpful 1 Interesting 2 Brilliant 0 Confused
Akshat Sharda
Aug 7, 2015

Its a G.P.(geometric progression)

Sum = a 1 − r =\frac{a}{1-r}

Where:

a a = first term ,i.e., ( 2 + 1 ) (\sqrt{2}+1)

r r = common ratio,i.e., ( 2 − 1 ) (\sqrt{2}-1)

Thus , 4 + 3 2 2 \boxed{\frac{4+3\sqrt{2}}{2}}

10 Helpful 0 Interesting 1 Brilliant 0 Confused

3 Helpful 0 Interesting 0 Brilliant 0 Confused

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