Algebra Pre-Algebra and Basic Algebra Math Forum

 January 9th, 2016, 06:23 AM #1 Newbie   Joined: Apr 2015 From: England Posts: 6 Thanks: 0 (Binomial Expansion) Could someone please help me? I have an exam next week and I have been practising some questions. Could anyone please give me the answers to the following questions? I was only able to do 6(a) and parts of 7. Would be really helpful. Thanks in advance. Last edited by skipjack; January 9th, 2016 at 06:45 AM.
 January 9th, 2016, 06:50 AM #2 Global Moderator   Joined: Dec 2006 Posts: 19,049 Thanks: 1618 Below are some hints or partial answers. 5. The expansion is $\displaystyle 2^6 + 6\left(2^5\right)Ax + 15\left(2^4\right)A^2x^2 + \,...$, so compare that to $64 - 576x + Bx^2$. 6.(a) $\displaystyle (1 - x)^7 = 1 - 7x + 21x^2 - 35x^3 + 35x^4 - \,...$ 6.(b) Use $x$ = 0.01 in the above. The last term shouldn't be needed. 7. $\displaystyle (1 + 4x)^4 = 1 + 16x + 96x^2 + 256x^3 + 256x^4$ and $\displaystyle (1 - 2x)^6 = 1 - 12x + 60x^2 - 160x^3 + 240x^4 - 192x^5 + 64x^6$. 8. Does $\displaystyle (1 + 2x)^n = 1 + 2nx + 2n(n - 1)x^2 + \,...$ help you? If you need further help, can you post what you were able to do, given the above?
 January 9th, 2016, 08:48 AM #3 Newbie   Joined: Apr 2015 From: England Posts: 6 Thanks: 0 5. I got as far as this: I failed to be able to find the values for a and b, I was completely stuck. 6. I was able to to this question, the answer that I got for b is: 7. I'm not sure if that's right, seems way too little for a 6 mark question. EDIT: I am now able to do it thanks to a friend who explained it to me. 8. completely confused, dont know where to even start @skipjack Thanks for the help in advance Last edited by skipjack; January 9th, 2016 at 11:30 AM.
 January 9th, 2016, 12:04 PM #4 Newbie   Joined: Apr 2015 From: England Posts: 6 Thanks: 0 EDIT: I am able to do all questions except 5 I got n=7 for number 8 if that's right Last edited by nautudent; January 9th, 2016 at 12:10 PM.
 January 9th, 2016, 12:07 PM #5 Global Moderator   Joined: Dec 2006 Posts: 19,049 Thanks: 1618 5. You got (Ax)², but changed it to Ax² instead of A²x². The next step should be to make the comparison I suggested. As 192A = -576, A = -576/192 = -3, and so B = 240A = -720. 6. You got the correct value. Do you see why only the first half of the complete expansion is needed? 7. The product includes (16x)(-12x) = -192x², so the answer is 156 - 192 = -36. 8. The coefficient of x² is six times the coefficient of x, so 2n(n - 1) = 6(2n), i.e. 2n² - 2n = 12n. Dividing by n (which is not zero) gives 2n - 2 = 12, so n = 7.

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