Please can anyone help me with this question..I'm a bit stuck. When (1-3/2x)^p is expanded in ascending powers of x, the coefficient of x is -24. a)find the value of p Many thanks, much appreciated
Please can anyone help me with this question..I'm a bit stuck. When (1-3/2x)^p is expanded in ascending powers of x, the coefficient of x is -24. a)find the value of p Many thanks, much appreciated
Using the binomial theorem for (1 + x)^n: (1 + x)^n = 1 + nx/1! + [n(n - 1)x^2]/2! + [n(n - 1)(n - 2)x^3]/3! + ...
Replace x with (-3/2)x, n = p
Alternatively, using the binomial theorem (a + b)^n = a^n + (nC1)a^(n-1)b + (nC2)a^(n-2)b^2 + … + (nCn-1)ab^(n-1) + b^n
Where (nCr) = n!/(r! (n - r)!), 1 = a, b = (-3x/2)
yes, so I got 1^p+(pC1)1^p-1(-3/2x)+(pC2)^p-2(-3/2x)^2...but im not sure how to expand.. how would I know what "r" to do it up to? (ie, in this case I done it up to 2)
Please can anyone help me with this question..I'm a bit stuck. When (1-3/2x)^p is expanded in ascending powers of x, the coefficient of x is -24. a)find the value of p Many thanks, much appreciated
yes, so I got 1^p+(pC1)1^p-1(-3/2x)+(pC2)^p-2(-3/2x)^2...but im not sure how to expand.. how would I know what "r" to do it up to? (ie, in this case I done it up to 2)
Looking at your question (quoted above), the coefficient of x is -24. Do you need to know what r is or do it all the way up to r?
btw: p!/(r!(n-r!) for (pC1) is p!/(1(p-1!). That means: [p(p-1)(p-2)(p-3)...(3)(2)(1)]/[1(p-1)(p-2)(p-3)....(3)(2)(1)]. If you cancel out the common factors from the numerators and denominators, it will make life a lot easier for yourself.
Also 1 to the power of anything doesn't change the answer
Looking at your question (quoted above), the coefficient of x is -24. Do you need to know what r is or do it all the way up to r?
btw: p!/(r!(n-r!) for (pC1) is p!/(1(p-1!). That means: [p(p-1)(p-2)(p-3)...(3)(2)(1)]/[1(p-1)(p-2)(p-3)....(3)(2)(1)]. If you cancel out the common factors from the numerators and denominators, it will make life a lot easier for yourself.
Also 1 to the power of anything doesn't change the answer