Adding up

Algebra Level 5

Find the value of 1/k


The answer is 1023.

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Karthik Sharma by Karthik Sharma , 22, India

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1 solution

Ayush Verma
Oct 30, 2014

If it comes in jee then put m=1,otherwise

l e t S = 1 r 2 r + 3 r 2 2 r + 7 r 2 3 r . . . . m t e r m s = ∑ n = 1 m T n w h e r e , T n = ( 2 n − 1 2 n ) r ∴ ∑ r = 0 10 ( − 1 ) r ( 10 r ) T n = ∑ r = 0 10 ( 10 r ) ( 1 − 2 n 2 n ) r = ( 1 + 1 − 2 n 2 n ) 10 = ( 1 2 10 ) n ∴ k ( 1 − 1 2 10 m ) = ∑ n = 1 m ( 1 2 10 ) n = 1 2 10 − 1 ( 1 − 1 2 10 m ) ⇒ 1 k = 2 10 − 1 = 1023 let\quad S=\cfrac { { 1 }^{ r } }{ { 2 }^{ r } } +\cfrac { { 3 }^{ r } }{ { 2 }^{ 2r } } +\cfrac { { 7 }^{ r } }{ { 2 }^{ 3r } } ....m\quad terms\\ \\ \quad \quad \quad =\sum _{ n=1 }^{ m }{ { T }_{ n } } \quad where,\\ \\ { T }_{ n }={ \left( \cfrac { { 2 }^{ n }-1 }{ { 2 }^{ n } } \right) }^{ r }\\ \\ \therefore \sum _{ r=0 }^{ 10 }{ { { (-1) }^{ r }\left( \begin{matrix} 10 \\ r \end{matrix} \right) } } { T }_{ n }=\sum _{ r=0 }^{ 10 }{ { \left( \begin{matrix} 10 \\ r \end{matrix} \right) } } { \left( \cfrac { { 1-2 }^{ n } }{ { 2 }^{ n } } \right) }^{ r }\\ \\ \quad \quad \quad \quad ={ \left( 1+\cfrac { { 1-2 }^{ n } }{ { 2 }^{ n } } \right) }^{ 10 }={ \left( \cfrac { 1 }{ { 2 }^{ 10 } } \right) }^{ n }\\ \\ \therefore k\left( 1-\cfrac { 1 }{ { 2 }^{ 10m } } \right) =\sum _{ n=1 }^{ m }{ { \left( \cfrac { 1 }{ { 2 }^{ 10 } } \right) }^{ n }= } \cfrac { 1 }{ { 2 }^{ 10 }-1 } \left( 1-\cfrac { 1 }{ { 2 }^{ 10m } } \right) \\ \\ \Rightarrow \cfrac { 1 }{ k } ={ 2 }^{ 10 }-1=1023

8 Helpful 0 Interesting 0 Brilliant 0 Confused

If you need to solve it in any competitive exam, and you are supposed to find the answer only. In that case put m = 1 m=1 and directly it will give you 1 k = 1023 \dfrac{1}{k}=1023 .

Sandeep Bhardwaj - 6 years, 7 months ago

I solved it same as your 2 method but i quite didnt understand how will m=1 give us the answer. Pls explain

Gautam Sharma - 6 years, 4 months ago

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