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