Almost: C(8,4)(2x)^4(-3)^4 90720x^4
ar^3 = 16
ar^6 = -128
r^3 = -8
r = -2
so, a = -2
a_n = -2*(-2)^(n-1) = -(-2)^n
S15 = 15/2 (2*8+14*4) = 540
a) use the binomial formula to find the 5th term of the expansion (2x - 3)^8 ------ i got 90720x?
b) Find a_n, the general term for a geometric sequence, given that the fourth term, a_4 is 16 and the seventh term is -128
c) Find the sum of 8+12+16+20+24+.....to 15 terms
thanks!
2 answers
D expression for (2x-3)^8 =(X)^8-8(X)^7(Y) 28(X)^6(Y)^2....70(X)^3(Y)^5... Put X=2x nd y=-3...there4 d 5th term =70(X)^3(Y)^5...=70*8x^3*-243=-136080..it seems the anxa z -136080.. Read more on pascal triangle or vandalmodes theorem
1 answer