⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ a 1 = 5 a n + 1 = 3 − a n 1 + 3 a n for n ≥ 2
A sequence { a n } satisfies the recurrence relation above. Find a 1 0 0 .
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.
Awesome solution, I don't even think of that! I use f ( x ) , and found that f 6 ( x ) = x .
Log in to reply
Done similar problem before, so I know the trick.
Log in to reply
Wow ,seems that you are familiar to many problems !
Log in to reply
@Kelvin Hong – I am on Brilliant every day. Problems solved 11,384. Current streak 1,183 Days. Solutions - 4074 total, ~26K upvotes. I think the answer given for #32 Hua Luo Geng is incorrect. I used an Excel spreadsheet and found that the answer is 1.43767E+90. Please check.
Log in to reply
@Chew-Seong Cheong – Wow ,I hope I can also solve problems in Brilliant everyday ! Besides that, the question #32 hua luo geng can turns into binomial expansion . Or may you write your solution to me? Maybe I have something wrong in the question.
Log in to reply
@Kelvin Hong – In that case, your answer (-1) is correct. I am a moderator, I can check the answer actually. I edited this problem actually.
Log in to reply
@Chew-Seong Cheong – Oh, have I given wrong information? I thought the expressions I have given isn't have any wrong .
Log in to reply
@Kelvin Hong – I have provided a solution for #32 and amended the problem wording
@Chew-Seong Cheong – Otherwise, your solution is actually same as mine.
Problem Loading...
Note Loading...
Set Loading...
Let a n = tan θ n , then we have, for n ≥ 2 ,
a n + 1 tan θ n + 1 = 3 − a n 1 + 3 a n = 3 − tan θ n 1 + 3 tan θ n = tan ( 3 π − θ n ) 1 = cot ( 3 π − θ n ) = tan ( 2 π − 3 π + θ n ) = tan ( 6 π + θ n )
⟹ θ n + 1 θ n ⟹ a 1 0 0 = 6 π + θ n = 6 ( n − 1 ) π + θ 1 = tan θ 1 0 0 = tan ( 6 ( 1 0 0 − 1 ) π + tan − 1 5 ) = tan ( 1 6 2 1 π + tan − 1 5 ) = tan ( 2 π + tan − 1 5 ) = − tan ( 2 π − tan − 1 5 ) = − cot ( tan − 1 5 ) = − 5 1 where θ 1 = tan − 1 5 Note that tan ( π − x ) = − tan x