Baller97 Posted May 11, 2015 Report Share Posted May 11, 2015 Hey guys If anybody could solve and explain this question to me, I'd really appreciate it. Given that (1+2/3x)n (3+nx)2 = 9 + 84x+..., find the value of n. Reply Link to post Share on other sites More sharing options...
Baller97 Posted May 11, 2015 Author Report Share Posted May 11, 2015 Anyone? Reply Link to post Share on other sites More sharing options...
Vioh Posted May 11, 2015 Report Share Posted May 11, 2015 Hey guys If anybody could solve and explain this question to me, I'd really appreciate it. Given that (1+2/3x)n (3+nx)2 = 9 + 84x+..., find the value of n. I got n = 7, is it the correct answer? Here's the working: By binomial theorem:[E1] Now, expanding (3+nx)^2, we have:[E2] Notice that equation [E2] doesn't have any 'k', so we can just combine that in the sum in equation [E1]:[E3] Expanding equation [E3] to get: [E4] From equation [E4], there are only 2 ways that you can get x^1. The first way is to let for the term Thus, The second way is to let for the term Thus, Now, adding the 2 ways together, we get: Hope this helps 1 Reply Link to post Share on other sites More sharing options...
Baller97 Posted May 11, 2015 Author Report Share Posted May 11, 2015 Hey guys If anybody could solve and explain this question to me, I'd really appreciate it. Given that (1+2/3x)n (3+nx)2 = 9 + 84x+..., find the value of n. I got n = 7, is it the correct answer? Here's the working: By binomial theorem:[E1] Now, expanding (3+nx)^2, we have:[E2] Notice that equation [E2] doesn't have any 'k', so we can just combine that in the sum in equation [E1]:[E3] Expanding equation [E3] to get: [E4] From equation [E4], there are only 2 ways that you can get x^1. The first way is to let for the term Thus, The second way is to let for the term Thus, Now, adding the 2 ways together, we get: Hope this helps Thank you so much!! I need your brain for the exam tomorrow Reply Link to post Share on other sites More sharing options...
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