<p>Any help on this would be great.</p>
<p>I just want to know shortcuts to solving combinations and sequence problems.</p>
<p>For instance, if the first term is 2 and to find the next term, you multiply the successive term by 5 and add 3. What is the 35th term?</p>
<p>What would I do to solve that?</p>
<p>bump......</p>
<p>you put the thing in your calc under npr i think</p>
<p>( 5 )
( 3 )</p>
<p>something like that... it's under "math" i think... cpr or npr.</p>
<p>to do that problem i would just do all 35 terms. a pain, i know, but you will get it right.</p>
<p>bump.... A fast way to do it</p>
<p>This one is a bit annoying. What you have to do is think of it as composite functions. f(x) = 5x+3, so you want g(f(x)) = 5(5x+3) + 3 etc. </p>
<p>You want to perform that 34 times right? Well if you start looking for patterns based on how many times you take the composite function you can see that you will always multiply x by 5, so you'll multiply by 5 34 times --- so your leading coefficient is 5^34 (times x) now for the adding it gets a little bit more tricky. </p>
<p>What basically happens is the first 3 will be multiplied by 5 34 times, the second 3 only 33 times, the third 32 times, etc. So you end up with 5^34<em>x + 5^34</em>3 + 5^33<em>3 + 5^32</em>3...etc etc. where x is 2 in this case. T</p>
<p>he general solution would be
5^(n-1)<em>x + 5^(n-1)</em>3 + 5^(n-2)<em>3...+ 5</em>3 + 3 where n is the term number.</p>
<p>I don't know of any way to simplify it from there. After looking at it this way I think you might be best just doing it with your calculator. Start by typing 2 [ENTER] and then typing 5*ANS + 3 and hitting enter enough times. That would be admittedly faster.</p>
<p>Btw x is your first term</p>
<p>As for a shortcut to permutations and combinations...I'm not sure what you're looking for here, but</p>
<p>The number of permutations of x number of choices is x! (or x * x-1 * x-2, etc.) 5! = 5<em>4</em>3<em>2</em>1 = 120. I.e. the number of words the can be formed from 8 distinct letters is 8! (since you're using them all.) If you were making only 4-letter words it would be 8<em>7</em>6*5 (8 choices for the first letter times 7 choices for the second and so on.)</p>
<p>The number of combinations of y things taken from a total of t things is
t! / ( y! * (t-y)!) </p>
<p>So, the number of sets of 3 marbles that can be taken from 10 marbles is
10! / (3! * (10-3)!) = 10! / (3! * 7!) = 10<em>9</em>8 / (3*2) = 120.</p>
<p>These are written 8 P 8, 8 P 4, and 10 C 3 respectively.
Hope that helps.</p>
<p>there is an easier way for that marble question...</p>
<p>10 nCr 3 = 120</p>
<p>nCr = second MATH... then hit the left key for PRB and then hit "3" for nCr.</p>
<p>ti-83 plus.</p>
<p>"For instance, if the first term is 2 and to find the next term, you multiply the successive term by 5 and add 3. What is the 35th term?"</p>
<p>Maybe I'm wrong, but I don't think that they'd ever give you a problem like that, because it combines an arithmetic sequence with a geometric sequence. The fastest way would be to use 5*Ans + 3 on your calculator as gmf suggested...</p>
<p>Memorize the formulas u learn in pre-cal regarding arithmetic and geometric sequences...</p>