Question

if k belongs to real no. and middle term of (k/2+2)^8 is 1120 .
find alue of k.

Answer 1

(k/2)+2 ??

Answer 2

1+8+28+56+70+56+28+8+1 then for a start right?

Answer 3

(k/2)^8
+8(2) (k/2)^7
+28(2)^2 (k/2)^6
+56(2)^3 (k/2)^5
+70(2)^4 (k/2)^4
+56(2)^5 (k/2)^3
+28(2)^6 (k/2)^2
+8(2)^7 (k/2)
+(2)^8

Answer 4

(k/2+2)=1120

Answer 5

the middle terms is:

70*2^4*(k/2)^4 = 1120

Answer 6

want to find value of k

Answer 7

70.16.k^4/16 = 1120

70.k^4 = 1120

k^4 = 1120/70

Answer 8

k = 4root(112/7) then

Answer 9

oo sorry (k/2+2)^8=1120

Answer 10

it said 'middle term' = 1120 ; you still want that?

Answer 11

ya

Answer 12

k = 4rt(16) = 2

k = +- 2; but id go with 2

Answer 13

ok.
thanks

Answer 14

70.16.(2^4/16) = 1120

70.16 = 1120

1120 = 1120 :)