Just put x = 1 x=1

Probability Level 5

( 1 − x 3 ) n = ∑ r = 0 n a r x r ( 1 − x ) 3 n − 2 r (1-x^3)^n = \sum_{r=0}^n a_r x^r (1-x)^{3n-2r}

Let a 1 , a 2 , … , a n a_1, a_2, \ldots , a_n be positive integer constants such that the above is an algebraic identity.

Suppose we set n = 5 n= 5 , find the value of ( ∑ k = 0 5 a k ) m o d 100 \displaystyle \left( \sum_{k=0}^5 a_k \right) \bmod {100} .


The answer is 24.

Rishi Sharma by Rishi Sharma , 32, India

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

1 solution

Rishi Sharma
Jan 15, 2017

We know, ( a 3 − b 3 ) = ( a − b ) ( a 2 + a b + b 2 ) \left( { a }^{ 3 }-{ b }^{ 3 } \right) =\left( a-b \right)\left( { a }^{ 2 }+ab+{ b }^{ 2 } \right) Now use a=1 & b=x to get ( 1 − x 3 ) = ( 1 − x ) ( 1 + x + x 2 ) (1-{ x }^{ 3 })=( 1-x )( 1+x+{ x }^{ 2 } ) So
∑ r = 0 n a r x r ( 1 − x ) 3 n − 2 r = [ ( 1 − x ) ( 1 + x + x 2 ) ] n \sum_{ r=0 }^{ n }{ { a }_{ r }{ x }^{ r }{ \left( 1-x \right) }^{ 3n-2r } } ={ \left[ \left( 1-x \right)\left( 1+x+{ x }^{ 2 } \right) \right] }^{ n } Dividing both sides by ( 1 − x ) 3 n \displaystyle { \left( 1-x \right) }^{ 3n } to get ∑ r = 0 n a r ( x 1 − 2 x + x 2 ) r = ( 1 + x + x 2 1 − 2 x + x 2 ) n ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ( i ) \sum_{ r=0 }^{ n }{ { a }_{ r }{ \left( \frac{ x }{ 1-2x+{ x }^{ 2 } } \right) }^{ r } } = { \left( \frac{ 1+x+{ x }^{ 2 } }{ 1-2x+{ x }^{ 2 } } \right) }^{ n } \cdot \cdot\cdot\cdot\cdot\cdot\cdot ( i ) We can notice that 1 + x + x 2 1 − 2 x + x 2 = 1 + 3 x 1 − 2 x + x 2 \displaystyle \frac{ 1+x+{ x }^{ 2 } }{ 1-2x+{ x }^{ 2 } } =1+ \frac{ 3x }{1-2x+{ x }^{ 2 } } . But ( 1 + 3 x 1 − 2 x + x 2 ) n = ∑ r = 0 n ( n r ) . 3 r ( x 1 − 2 x + x 2 ) r { \left( 1+\frac{ 3x }{ 1-2x+{ x }^{ 2 } } \right) }^{ n } = \sum_{r=0 }^{ n } { { n \choose r } . { 3 }^{ r }{ \left( \frac{ x }{1-2x+{ x }^{ 2 } } \right) }^{ r } } Comparing it with ( i ) (i) . we get a r = ( n r ) 3 r { a }_{ r } = { n \choose r }{ 3 }^{ r } Hence, ∑ k = 0 5 a k = ∑ k = 0 5 ( 5 k ) 3 k = 4 5 = 1024 \displaystyle \sum_{ k=0 }^{ 5 }{ { a }_{ k } }=\sum_{ k=0 }^{ 5 }{ { 5 \choose k }{ 3 }^{ k } } = { 4 }^{ 5 } = \boxed{ 1024 }

4 Helpful 0 Interesting 0 Brilliant 0 Confused

Instead in ( i ) (i) you can set x 1 − 2 x + x 2 = 1 \frac{ x }{ 1-2x+{ x }^{ 2 } }=1 or x + 1 x = 3 x+\frac 1x=3 . So LHS of i i becomes required expression while R H S RHS is simply ( x 2 + x + 1 1 − 2 x + x 2 ⏟ x ) n = ( x + 1 x + 1 ) n = 4 n \left(\frac{ x^2+x+1 }{ \underbrace{1-2x+{ x }^{ 2 }}_x } \right)^n=\left(x+\frac 1x +1\right)^n=4^n and put n=5.

Btw nice question...

Rishabh Jain - 4 years, 4 months ago

Log in to reply

Nice observation.

Rishi Sharma - 4 years, 4 months ago

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...