Binomial Newton

Probability Level 4

Given f ( x ) f(x) denotes the number of consecutive zeroes from the right. As an example, f ( 20100 ) = 2 , f ( 1000 ) = 3 , f ( 2018 ) = 0 f(20100) = 2, f(1000)=3, f(2018)=0 . Determine the value of f ( ∑ j = 0 n ( ( n j ) ( ∑ i = 0 j ( j i ) 998 i ) ) ) n \frac {\displaystyle f\left( \sum _{ j=0 }^{ n }{ \left( \left( \begin{matrix} n \\ j \end{matrix} \right) \left( \sum _{ i=0 }^{ j }{ \left( \begin{matrix} j \\ i \end{matrix} \right) } { 998 }^{ i } \right) \right) } \right) }{ n }


The answer is 3.

Wildan Bagus Wicaksono by Wildan Bagus Wicaksono , 18, Indonesia

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1 solution

Note that, ( x + 1 ) n = ∑ i = 0 n ( n i ) x i \displaystyle { (x+1) }^{ n }=\sum _{ i=0 }^{ n }{ \left( \begin{matrix} n \\ i \end{matrix} \right) } { x }^{ i } .

∑ i = 0 j ( j i ) 998 i = ( 998 + 1 ) j ∑ i = 0 j ( j i ) 998 i = 999 j ∑ j = 0 n ( ( n j ) ( ∑ i = 0 j ( j i ) 998 i ) ) = ∑ j = 0 n ( n j ) 99 9 i ∑ j = 0 n ( ( n j ) ( ∑ i = 0 j ( j i ) 998 i ) ) = ( 999 + 1 ) n ∑ j = 0 n ( ( n j ) ( ∑ i = 0 j ( j i ) 998 i ) ) = 1000 n ∑ j = 0 n ( ( n j ) ( ∑ i = 0 j ( j i ) 998 i ) ) = 10 3 n f ( ∑ j = 0 n ( ( n j ) ( ∑ i = 0 j ( j i ) 998 i ) ) ) = 3 n \displaystyle \sum _{ i=0 }^{ j }{ \left( \begin{matrix} j \\ i \end{matrix} \right) } { 998 }^{ i }={ (998+1) }^{ j }\\ \displaystyle \sum _{ i=0 }^{ j }{ \left( \begin{matrix} j \\ i \end{matrix} \right) } { 998 }^{ i }={ 999 }^{ j }\\ \displaystyle \sum _{ j=0 }^{ n }{ \left( \left( \begin{matrix} n \\ j \end{matrix} \right) \left( \sum _{ i=0 }^{ j }{ \left( \begin{matrix} j \\ i \end{matrix} \right) } { 998 }^{ i } \right) \right) } = \displaystyle \sum _{ j=0 }^{ n }{ \left( \begin{matrix} n \\ j \end{matrix} \right) { 9 }99^{ i } } \\ \displaystyle \sum _{ j=0 }^{ n }{ \left( \left( \begin{matrix} n \\ j \end{matrix} \right) \left( \sum _{ i=0 }^{ j }{ \left( \begin{matrix} j \\ i \end{matrix} \right) } { 998 }^{ i } \right) \right) } ={ (999+1) }^{ n }\\ \displaystyle \sum _{ j=0 }^{ n }{ \left( \left( \begin{matrix} n \\ j \end{matrix} \right) \left( \sum _{ i=0 }^{ j }{ \left( \begin{matrix} j \\ i \end{matrix} \right) } { 998 }^{ i } \right) \right) } ={ 1000 }^{ n }\\ \displaystyle \sum _{ j=0 }^{ n }{ \left( \left( \begin{matrix} n \\ j \end{matrix} \right) \left( \sum _{ i=0 }^{ j }{ \left( \begin{matrix} j \\ i \end{matrix} \right) } { 998 }^{ i } \right) \right) } ={ 10 }^{ 3n }\\ \displaystyle f\left( \sum _{ j=0 }^{ n }{ \left( \left( \begin{matrix} n \\ j \end{matrix} \right) \left( \sum _{ i=0 }^{ j }{ \left( \begin{matrix} j \\ i \end{matrix} \right) } { 998 }^{ i } \right) \right) } \right) =3n

So, we get

f ( ∑ j = 0 n ( ( n j ) ( ∑ i = 0 j ( j i ) 998 i ) ) ) n = 3 n n ∴ f ( ∑ j = 0 n ( ( n j ) ( ∑ i = 0 j ( j i ) 998 i ) ) ) n = 3 \frac { \displaystyle f\left( \sum _{ j=0 }^{ n }{ \left( \left( \begin{matrix} n \\ j \end{matrix} \right) \left( \sum _{ i=0 }^{ j }{ \left( \begin{matrix} j \\ i \end{matrix} \right) } { 998 }^{ i } \right) \right) } \right) }{ n } =\frac { 3n }{ n } \\ \therefore \frac { \displaystyle f\left( \sum _{ j=0 }^{ n }{ \left( \left( \begin{matrix} n \\ j \end{matrix} \right) \left( \sum _{ i=0 }^{ j }{ \left( \begin{matrix} j \\ i \end{matrix} \right) } { 998 }^{ i } \right) \right) } \right) }{ n } =3

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