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use pascal's triangle for this one. go down to the 11th row of the triangle and find the third term
I think I am having trouble setting up pascals triangle for this one. Can you help me? I have a few problems like this
if you see 6C3 , do you know what it mean?
no should I?
1 11 121 1331 14641..... and keep doing this for 11 rows
have to, search google about pascal and something related, because pascal triangle is good if the exponent is low, as 3,4,5, to 11 and up, pascal takes you forever to solve
however, pascal is basic, ok, let dubindabay help you
why do you use numbers if it says (a+b)^11? Another problem I have is (x+2)^5
pascal's triangle tells you the coefficients of each term. so in the general form you have something that looks a little like this: p_1*a^n*b^0+p_2*a^n-1*b^1..... so the first term of your problem would be: 1*a^11 the second term is: 11*a^10*b
I wish I could understand. Its not clicking.
can I say something, @dubindabay
of course @Loser66
so, pascal triangle is the number you can "PUT" in the front of the terms in exponent polynomial. got what I mean?
you got it!
oh my bad I confused you with the asker ;P
Thankyou for the effort everyone, I will try my best to solve this problem
what do you mean? give up?
we are here, try to help, but if you give up, I have no reason to NOT give up, hehehe... how about dubindabay?
So should I just write the triangle out, all 11 rows? 1 11 121 1331 14641 1_5_10_10_5_1 1_6_15_20_15_6_1 1_7_21_35_35_21_7_1 1_8_28_56_70_56_28_8_1 1_9_36_84_126_126_84_36_9_1 1_10_45_120_210_252_210_120_45_10_1 so the answer is 45?
that 's just the coefficient of the third term, the terms of that always have 3 parts number (coefficient) a ^(something) b^(something) try to figure out those something and put them under the form of (45 a^...b^...)
look at my box above, a^2 +2 ab + b^2, don't be confused because it actually is |dw:1370560200616:dw|
why is the highest for the b exponent 1, not 2?
yes, my bad, I am sorry, mistake, it's 2
So for my problem (a+b)^11, how do i find the third term? I am having trouble visualizing it. All I have is 1 11 121 1331 14641 1_5_10_10_5_1 1_6_15_20_15_6_1 1_7_21_35_35_21_7_1 1_8_28_56_70_56_28_8_1 1_9_36_84_126_126_84_36_9_1 1_10_45_120_210_252_210_120_45_10_1 Where do i go from here?
based on the triangle, what is the 3rd term of the 11th row?
(you said it earlier)
I am stuck trying to figure out 45a^(something)b^(something) + ... + ...
good. that is the coefficient of the third term. then you need two more pieces. in the first term, you have a coefficient of 1, and a^11, and b^0. For the second term, you have a coefficient of 10, a^10, and b^1.
I'll skip the third term to let you figure it out. but just for reference, the fourth term is: 120a^8b^3
45a^9b^2? I am just following your pattern, i dont know how to do this if this is correct. Where did you get 1, a^11 and b^0 for the first term?
is it a^11 because there is 11 terms in the 11th row?
correct!! it is just a pattern.
its a^11 because your raising the function to the 11th power there are 11 rows because its the 11th power
why do you start at b^0?
is it because 0 is the lowest number? From a demonstration above he said that it goes exponent from highest to lowest for a and lowest to highest for b. Why is 0 the lowest exponent? 0-11 would give 12 terms, and you use 11 as the highest exponent.
im basically quoting this good. that is the coefficient of the third term. then you need two more pieces. in the first term, you have a coefficient of 1, and a^11, and b^0. For the second term, you have a coefficient of 10, a^10, and b^1.
he says a^11 for the first power, and b^0 for the last power. That would mean there are 12 terms right? 11, 10,9,8,7,6,5,4,3,2,1,0? So instead shouldnt it be 1a^10b^0
very good question. there are actually 12 terms in (a+b)^11 look at pascals triangle. there are 12 numbers in the 11th row
1 11 121 1331 14641 1_5_10_10_5_1 1_6_15_20_15_6_1 1_7_21_35_35_21_7_1 1_8_28_56_70_56_28_8_1 1_9_36_84_126_126_84_36_9_1 1_10_45_120_210_252_210_120_45_10_1
did i draw my triangle wrong? I only count 11 terms in the 11th row
yeah I should've caught that. the third term is actually 55 a^9b^2
theres a small mistake somewhere. but for example, if you take (a+b)^2 how many terms will you have?
im lost again :( I think I have a very wrong understanding of this. my original question was What is the third term of (a + b)^11? So your saying the answer to that question is : 55a^9b^2 and so I messed up somewhere in the triangle because I thought it was 45a9b2 Now i dont know how many terms I will have if I take (a+b)^2 :(
"foil" the (a+b)^2. (a+b)*(a+b)=a^2+2ab+b^2
I count 3 terms. which is one more than the power that (a+b) is raised to! :)
so back to my problem. Can you explain this problem to me. Every time I try to figure it out I get more confused. What is the third term of (a + b)^11?
I thought I was starting to understand but I am back at square one again :(
don't worry, you're making progress. you had some very good questions. before we do your problem, let me ask you one more question. what is (a+b)^3?
(a+b)(a+b)(a+b) (a^2 + ab + ab + b^2)(a + b) a^3 + a^2b + a^2b + b^2a + a^2b + b^2a + b^2a + b^3 Am I just totally clueless?
heres another pascals triangle: 1 ----------- corresponds to raising (a+b) to the 0th power 1 1----------1st power 1 2 1---------2nd power 1 3 3 1--------3rd power 1 4 6 4 1------4th power 1 5 10 10 5 1--5th power 1 6 15 20 15 6 1-6th power 1 7 21 35 35 21 7 1-7th power
no thats absolutely correct. now just add your like terms!
Thankyou for the correct drawing of the pascals triangle. I thought the first row starting at 1st power. So I am missing a row then here? 1 11 121 1331 14641 1_5_10_10_5_1 1_6_15_20_15_6_1 1_7_21_35_35_21_7_1 1_8_28_56_70_56_28_8_1 1_9_36_84_126_126_84_36_9_1 1_10_45_120_210_252_210_120_45_10_1 Should there be one more row for my problem What is the third term of (a + b)^11?
that's it! :)
I missed that too
so the correct answer is still 55 a^9b^2? And the max value for the powers is 11 because there is 12 terms in the 12th row? And the minimum value for the powers is always 0 no matter what? And because 55 is the third term the answer is 55 a^9b^2 because a^11b^0 is for the first term, a^10b^1 is for the 2nd term and a^9b^2 is the pattern for the third term Thus the max power is always 1 less than the number of terms in the row. And the number of terms in the row is equal to the row number. 12 terms being in the 12th row which is the row of the 11th power?
yep, sounds like you've got it down!
its easier to say that the max power is going to be 11 because the function is raised to the 11th power, but other than that everything you said sounds good.
Thankyou for your time. To think this is only my first problem ! But I am very happy you stayed with me through it.
now im so used to the question being (a+b)^ something power. What do I do if the inside of the () changes? Use Pascal's triangle to expand (x + 2)^5.
the time you spent on this one to really understand it will make the other ones much easier. It was a pleasure to help
(x+2)(x+2)(x+2)(x+2)(x+2) That is not what the problem is asking is it?
now you have a general form for (a+b)^something. so in the case of (x+2)^5, all you have to do is follow the same format, just substitute x for a and 2 for b
thats exactly what the problem is asking. but with pascal's triangle it's much easier to find the solution
So can I use your triangle you gave me for (a+b)^11? Though I know this only goes to the 6th row unlike the 12th row for the 11th power. 1 ----------- corresponds to raising (a+b) to the 0th power 1 1----------1st power 1 2 1---------2nd power 1 3 3 1--------3rd power 1 4 6 4 1------4th power 1 5 10 10 5 1--5th power
What do you have to do differently because inside the () is (x + 2) not (a+b) like last time?
the only difference is that now you have x instead of "a" and 2 instead of "b". other than that, you use the exact same pattern
so my problem is Use Pascal's triangle to expand (x + 2)^5. Is the answer to this problem 1 ----------- corresponds to raising (a+b) to the 0th power 1 1----------1st power 1 2 1---------2nd power 1 3 3 1--------3rd power 1 4 6 4 1------4th power 1 5 10 10 5 1--5th power ?
how do incorporate the (x+2) as I am sure it has to be involved in the answer. Is it x^5 + 2^0
I'll do (x+2)^4 for you. we follow the same idea as in (a+b)^4 1 ----------- corresponds to raising (a+b) to the 0th power 1 1----------1st power 1 2 1---------2nd power 1 3 3 1--------3rd power 1 4 6 4 1------4th power <------this is the one we want to use for (a+b)^4 we have: \[a^4+4 a^3 b+6 a^2 b^2+4 a b^3+b^4\] substituting x-->a and 2-->b we get \[(x^4*2^0)+(4x^3*2^1)+(6x^2*2^2)+(4x^1*2^3)+(x^0*2^4)\] we can simplify this to \[x^4+8x^3+24x^2+32x+16\]