differentiation question (1 Viewer)

CM_Tutor

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Rewrite using log laws - ie (sin x)<sup>cos x</sup> = e<sup>ln(sin x)<sup>cos x</sup></sup> = e<sup>cos x * ln(sin x)</sup>
 

untamedanimal

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This is on a different topic but seeing that i started this thread i might aswell post it here. Ive been doing binomials on my own before we learn it at school and ive come stuck on two questions in the Fitzpatrick book. They may not be that hard

They are from Exercise 29 (a)

40. Find the value of r if the coefficients of the rth term from the beginning and the rth term from the end of (2x + 3)^15 are in the ratio 8:27

43. The 2nd, 3rd and 4th terms in the expansion of (a + b)^n are 12, 60 and 160 respectively. Find the values of a, b and n
 

CM_Tutor

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Q43: (a + b)<sup>n</sup> = <sup>n</sup>C<sub>0</sub>a<sup>n</sup> + <sup>n</sup>C<sub>1</sub>a<sup>n-1</sup>b + <sup>n</sup>C<sub>2</sub>a<sup>n-2</sup>b<sup>2</sup> + <sup>n</sup>C<sub>3</sub>a<sup>n-3</sup>b<sup>3</sup> + ... + <sup>n</sup>C<sub>n</sub>b<sup>n</sup>

Now, we know that the second, third and fourth terms are 12, 60 and 160. It follows that:
<sup>n</sup>C<sub>1</sub>a<sup>n-1</sup>b = na<sup>n-1</sup>b = 12 _____ (1)
<sup>n</sup>C<sub>2</sub>a<sup>n-2</sup>b<sup>2</sup> = n(n - 1)a<sup>n-2</sup>b<sup>2</sup> / 2 = 60 _____ (2)
<sup>n</sup>C<sub>3</sub>a<sup>n-3</sup>b<sup>3</sup> = n(n - 1)(n - 2)a<sup>n-3</sup>b<sup>3</sup> / 6 = 160 _____ (3)

Solve these three simultaneous equations for a, b and n

Q40: Expand (2x + 3)<sup>15</sup> just as I expanded (a + b)<sup>n</sup>, above, and identify the r<sup>th</sup> terms from each end, and put the coefficients into the equation Coeff r<sup>th</sup> term from start / Coeff r<sup>th</sup> term from end = 8 / 27
 

Xayma

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Copy and paste ;), or just learn to write the < sup > and < sub > tags quickly.
 

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