An algebra problem by I Gede Arya Raditya Parameswara

Algebra Level 4

f ( n + 2 ) + 7 f ( n + 1 ) + 10 f ( n ) = 0 f(n+2)+7f(n+1)+10f(n)=0 f ( 0 ) = 2 f(0)=2 f ( 1 ) = − 13 f(1)=-13

If f ( 100 ) f(100) can be expressed as a ( b − 3 ) 100 − c ( 102 − d ) 100 a(b-3)^{100}-c(102-d)^{100} where a > 0 , b − 3 > 0 , 102 − d > 0 a> 0, b-3>0 , 102-d>0 and c < 101 c<101 , then what is the value of ( b + c + d ) a (b+c+d)^{a} ?


The answer is 1295029.

I Gede Arya Raditya Parameswara by I Gede Arya Raditya Para… , 17

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1 solution

The formula is f ( n ) = 3 ( − 5 ) n − ( − 2 ) n f(n)=3(-5)^{n}-(-2)^{n} Because 100 is even and b − 3 > 0 ] a n d , \[ 102 − d > 0 b-3>0]\ and, \[102-d>0 so we can write f ( 100 ) = 3 ( 5 ) 100 − 2 100 f(100)=3(5)^{100}-2^{100} So, ( b + c + d ) a = 1295029 (b+c+d)^{a}=1295029

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@Dipa Parameswara Can you explain how you got that formula ..!!??Please!!

Ankit Kumar Jain - 4 years, 3 months ago

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binet's formula

I Gede Arya Raditya Parameswara - 3 years, 5 months ago

@Jon Haussmann Sir , can you also please explain the method , please!!?

Ankit Kumar Jain - 4 years, 3 months ago

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Thanks a lot ..!!

Ankit Kumar Jain - 4 years, 3 months ago

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