Expert: Ahmed Salami - 10/6/2004

Question
Find the value of r if the coefficients of the rth term from the begining and the rth term from the end of the expansion of (3x+5)^15 are in the ratio 27:125.



Can you just explain to me what the question is asking about.Thanks

Answer
Hi Hairani,
For the expansion of (3x+15)^15, the coefficients of the rth term from the beginning and end are
15C(r-1).(3x)^16-r.(5)^r-1 and
15C(16-r).(3x)^r-1.(5)^16-r
But the binomial coefficients 15C(r-1)and 15C(16-r)are equal and so we leave them out of the consideration. Considering the coefficients of x(while ignoring the power), they are (3)^16-r.(5)^r-1 and (3)^r-1.(5)^16-r
The ratio is therefore (3)^16-r.(5)^r-1/(3)^r-1.(5)^16-r
= 3^17-2r. 5^2r-15 = (3/5)^17-2r
But (3/5)^17-2r = 27/125 = (3/5)^3
So,
17-2r = 3
2r = 14
r = 7

I hope you understand it.
Regards.