binomial theorem help? What's the 12th term? ?
Determine the exact value of the 12th term in the expansion of (x^2 - 1/2)^16
Determine the exact value of the 12th term in the expansion of (x^2 - 1/2)^16
Engr. Ronald
(x^2 - 1/2)^16
=x^32 - 8 x^30 + 30 x^28 - 70 x^26 + (455 x^24)/4 - (273 x^22)/2 + (1001 x^20)/8 - (715 x^18)/8 + (6435 x^16)/128 - (715 x^14)/32 + (1001 x^12)/128 - (273 x^10)/128 + (455 x^8)/1024 - (35 x^6)/512 + (15 x^4)/2048 - x^2/2048 + 1/65536
We see the 12th term is (- 273x^10)/128..... Answer//
Alan
for generic (x+y)^n
kth term = ( n (k-1) ) x^((n- (k-1) ) y^((k-1)
12th term
n = 16
k = 12
k -1 = 11
(16 11) x^ (16 -11) y^11
(16 11) x^5 y^11
16! / (11! 5!) = 16*15*14*13*12 / 5*4*3*2*1
since 15 = 5*3*1
and 16/(4*2) = 2
(16 11) = 2*14*13*12 = 4368
now substitute
x^2 for x
and
(-1/2) for y
4368 *( x^2) ^5 (-1/2)^11
4368 *(x^10) (-1)^11 / (2^11)
(-4368/2048) *x^10
Answer:
-(273/128) x^10
or
(-2 17/128)x^10
Krishnamurthy
(x^2 - 1/2)^16
= x^32 - 8 x^30 + 30 x^28 - 70 x^26 + (455 x^24)/4 - (273 x^22)/2 + (1001 x^20)/8 - (715 x^18)/8 + (6435 x^16)/128 - (715 x^14)/32 + (1001 x^12)/128 - (273 x^10)/128 + (455 x^8)/1024 - (35 x^6)/512 + (15 x^4)/2048 - x^2/2048 + 1/65536
The 12th term is - (273 x^10)/128