<p>I found this problem and need help with it. Thanks!!
The first three terms of a sequence are 1/(1)(2), 1/(2)(3), and 1/(3)(4). The nth term of the sequence is 1/(n)(n+1), which is equal to 1/n - 1/(n+1). What is the sum of the first 50 terms of this sequence?
a) 1
b) 50/51
c) 49/50
d) 24/50
e) 1/(50)(51)</p>
<p>Is the answer B?</p>
<p>yes it is…</p>
<p>Well I’m really bad at explanations so I’m not much help. I’m pretty sure another CCer will explain</p>
<p>That type of question is not tested on the SAT.</p>
<p>Actually I’ll give it a try. All I did was find a pattern to get my answer. I realized that 1/2 or 1/(1)(2) plus 1/6 or 1/(2)(3) was 2/3, and 1/6 was the second term of the sequence. So if you were to keep adding until 50 you would get 50/51. </p>
<p>… I just re read what I wrote and I would understand if you coined this as the worst explanation ever lol</p>
<p>@dusterbug- It actually is… one of the harder ones I have seen
@Vapperss- haha… I think I know what u r saying… so u just brought the 50 on top, just like u put the 2 on top and the 3 on the bottom for the sum of 1/(1)(2) and 1/(2)(3)? If so, I wonder if there is another way… that requires alot of guess work haha</p>
<p>Well I didn’t just bring it to the top, I added the third term to my total to see if there really was in fact a pattern. Sure enough I got 3/4, so that’s when I knew that if I kept going to 50 then it would be 50/51.</p>
<p>*btw congrate on your 400th post! Woot woot</p>
<p>The “hint” at the end of the problem statement is meant to tell you how to proceed.</p>
<p>1/[(n)(n+1)] = 1/n - 1/(n+1)</p>
<p>So the Sum 1/[(j)(j+1)] from 1 to 50 is the Sum of [ 1/j - 1/(j+1)] from 1 to 50.</p>
<p>The Sum [1/(j+1)] from 1 to 50 is equal to the Sum [1/j] from 1 to 50 + 1/51 - 1. (Write out a few terms to convince yourself). The Sum [1/j] terms cancel out and you’re left with 1 - 1/51.</p>
<p>Try it out where the sum is from 1 to 75 or from 1 to 100.</p>
<p>Use mathematical induction to find and be sure of summation formula, i.e.,</p>
<p>Assume that 1/(1<em>2) + 1/(2</em>3) + … + 1 / (n*(n+1)) = n/(n+1)</p>
<p>Then </p>
<p>[ 1/(1<em>2) + 1/(2</em>3) + … + 1 / (n<em>(n+1)) ] + 1 / ((n+1)</em>(n+2)) </p>
<p>= n/(n+1) + 1 / ((n+1)*(n+2)) — because of what we assumed</p>
<p>= n(n+2) / ((n+1)<em>(n+2)) + 1 / ((n+1)</em>(n+2)) — to get common denominator</p>
<p>= ((n^2)+2n+1) / ((n+1)*(n+2)) — combining terms</p>
<p>= ((n+1)^2) / ((n+1)*(n+2)) — factoring</p>
<p>= (n+1) / (n+2)</p>
<p>Thus, we complete the proof by showing that </p>
<p>if 1/(1<em>2) + 1/(2</em>3) + … + 1 / (n*(n+1)) = n/(n+1)</p>
<p>then the next term will sum to (n+1) / (n+2)</p>
<p>Applying the summation formula, </p>
<p>n/(n+1) = 50/51 for term fifty</p>
<p>(b) 50/51 </p>
<p>We guessed at the formula by summing the first three sets independently to </p>
<p>1/2, 2/3, 3/4</p>
<p>1st term - 1/1 - 1/2
2nd term - 1/2 - 1/3
.
.
49th term - 1/49 - 1/50
50th term - 1/50 - 1/51</p>
<p>Sum = (1/1 - 1/2) + (1/2 -1/3) + (1/3 - 1/4) + … + (1/49 - 1/50) + (1/50 - 1/51)
1st term 2nd term 3rd term 49th term 50th term</p>
<p>Opening the parentheses,</p>
<p>Sum = 1/1 -1/2 +1/2 -1/3 + 1/3 -1/4 +…+ 1/49 - 1/50 +1/50 -1/51</p>
<p>All except the first and last fractions cancel out due to +ve and -ve terms. Then,</p>
<p>Sum = 1/1 - 1/51 = 50/51</p>
<p>Thanks guys!! I understand it now. I do have a couple more… so any help would be appreciated!
- In the triangle ABC, AB=6 and AC=8. If x>90, what is one possible length of BC (which is opposite of X).
How do I solve this? - In the triangle ABC, AB=AC=5 and BC=6. What is the area of the triangle?
a)4
b)6
c)9
d)12
e)16 - The integers 1 through 6 appear on the six faces of a cube, one on each face. If three such cubes are rolled, what is the probability that the sum of the numbers on top faces is 17 and 18?
a) 1/108
b) 1/54
c)1/27
d)1/18
e)1/16
THANKS!!!</p>
<p>For (3) – There is only one way the sum can be 18, specifically when each of top faces is 6. The probability of that happening is (1/6)<em>(1/6)</em>(1/6)=1/216. There are three ways that the sum can be 17, specifically when 2 of the faces are 6 and 1 face is 5. The probability of that is 3<em>(1/6)</em>(1/6)*(1/6)=3/216. Adding 1/216 + 3/216 = 4/216 = 1/54 – i.e choice (b).</p>
<p>For (2) … The triangle is isosceles. The base is BC = 6. Draw a line from A perpendicular to BC. Assume that it intersects BC at point P. The length of BP=PC=3. AB (=5) is the hypotenuse of the triangle ABP. The height is therefore 4 (it’s a 3-4-5 triangle). Area is Base*Height/2 or 12.</p>
<p>For (1) … Is x the angle BAC? If x=90 degrees then X=BC=10 (The triangle is twice a 3-4-5 triangle … or apply Pythagoras theorem). If x is greater than 90 than BC is greater than 10. You don’t show the choices, so I can’t “pick” an answer.</p>
<p>Also, BC is less than 14. Because that is the sum of other two sides. So, 10<BC<14.</p>
<p>thanks a ton guys!! Those were really easy ones lol. I just wasnt thinking.
and @fogcity- The reason I didnt give answer choices was because it was one of the grid in ones :)</p>
<p>… I have one more (for now lol) if u guys want to help…
1)The lengths of the sides of an isosceles triangle are 20, n, and n. If n is an integer, what is the smallest perimeter for the triangle?
a)40
b)41
c)42
d)44
e)60</p>
<p>I think it would be: c) 42
In a triangle, the minimum sum of two sides should be greater than the third side.
It cannot be 40 because 20+2n = 40 would make 2n=20
It cannot be 41 because 20+2n = 41 would make n = 10.5 & n has to be an integer
For the perimeter to be 42 to be is possible with n = 11 & that would be the smallest perimeter.</p>
<p>ur absolutely correct! Thanks… I keep forgetting about these rules I learned years ago haha. BC calc doesnt help too much with simple SAT problems lol</p>