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

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

Mathematics
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(k/2)+2 ??
1+8+28+56+70+56+28+8+1 then for a start right?
(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

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Other answers:

(k/2+2)=1120
the middle terms is: 70*2^4*(k/2)^4 = 1120
want to find value of k
70.16.k^4/16 = 1120 70.k^4 = 1120 k^4 = 1120/70
k = 4root(112/7) then
oo sorry (k/2+2)^8=1120
it said 'middle term' = 1120 ; you still want that?
ya
k = 4rt(16) = 2 k = +- 2; but id go with 2
ok. thanks
70.16.(2^4/16) = 1120 70.16 = 1120 1120 = 1120 :)

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