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Post Info TOPIC: it's quincy...i am back, of course...with questions


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it's quincy...i am back, of course...with questions
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hey everyone,
...yes i have to admit i've been super super super lazy on posting...lol, because i've been practicing so hard...yes i have been...and i ran across some questions on column 4 that i can't solve....yeah...i am finally beginning to finish tests...i have about 2:30 left when i get to the 4th column but...i dunno the tricks to some of them...and that bites...cuz it slows me down alot...here it goes:
1. 306^2+603^2= (...tried to derive it, thought i did derive it, nope, i was wrong)
2. 14*15/16-17= (...i'll work this some more...the trick looks...derivable)
3. 1/28+1/36+1/45= (can't see what the relationship is)
4. the sum of the first seven terms in the fibonacci series is....(yeah, i've been dying to know the formula for this one)
thanks thanks thanks guys...hopefully yall are alive...becuz no one really posts anymore here...lol, laziness?

yeah...i basically made the same post of texasmath...just to see who answers first...cuz it seems like everyone is hybernating....

quincy

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Anonymous

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1. Well, you know 306^2 is 93636, and 603^2 is 363609, so add them up, and you get 457245.  OR, since you know the answer will be "symmetrical", find the last digits of 306^2 (36), add it up to the first digits of 306^2 (9). So your answer is 45__45.  you know (a+b)^2 = a^2 + 2ab+b^2, the middle is 4ab or 4*3*6 = 72. 457245.  I hope that makes sense, it's all clear in my head, but not so much on paper.


 


2. I don't know that there's really a trick, so just work it out.  THe 14 and 16 reduce to 7/8 * 15. or 105/8.  That's 13 1/8, - 17 is -3 7/8.  Not too long,


3. Consecutive Triangular numbers in the denominator means 3/(2nd denominator - 1). 3/35


4. F(n+2) -1, or F(9)-1. Means 34-1 = 33.


Hope this helps!


 



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yeah, it helped alot except for the fibonacci sequence...to this very day, after 314159265359 ppl have asked the same question, i still think there is an easy way for finding the sum of fibonacci numbers (instead of memorizing)...or do F(n+2) -1. becuz we'll have memorize n to about the 13th or 14th fib number....but since i can't come up with a working formula...i'll stick with the F(n+2) -1 formula and start memorizing fib. numbers over the break...

thanks

quincy

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Anonymous

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On All the Number Sense Tests I've seen so far, they take the fibonacci sum for 10 terms.


For that, multiply the 7th term by 11.


It wouldn't work for your question though.


Vinay



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So it would be like this?

x+y+(x+y)+(x+2y)+(2x+3y)+(3x+5y)+(5x+8y)+(8x+13y)+(13x+21y)+(21x+34y)=55x+88y=11(5x+8y), which is the 7th term....cool.

Does it work only for the first ten terms?

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Anonymous

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looks like it... but remember, it'll work for any sort of sequence, even if it's not fibonacci or doesn't begin with one. that's kinda cool to me.



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