Question

N-натуральное число F(1)=1 G(1)=1 F(n)=2*f(n-1)-g(n-1) G(n)=F(n-1)+2*g(n-1) G(5)-f(5)=?

Answer 1

F(1)=1; g(1)=1
f(2)=2*f(1)-g(1)=2-1=1
g(2)=f(1)+2*g(1)=1+2=3
f(3)=2*f(2)-g(2)=2*1-3=-1
g(3)=f(2)+2*g(2)=1+2*3=7
f(4)=2*f(3)-g(3)=2*(-1)-7=-9
g(4)=f(3)+2*g(3)=-1+2*7=13
f(5)=2*f(4)-g(4)=2*(-9)-13=-31
g(5)=f(4)+2*g(4)=-9+2*13=17
g(5)-f(5)=16-(-31)=16+31=47
Ответ: 47

Answer 2

To find the value of G(5) - F(5), we'll first calculate the values of F(n) and G(n) for n = 2, 3, 4, and 5 using the given recursive formulas:

For n = 2: F(2) = 2 * F(1) - G(1) = 2 * 1 - 1 = 1 G(2) = F(1) + 2 * G(1) = 1 + 2 * 1 = 3

For n = 3: F(3) = 2 * F(2) - G(2) = 2 * 1 - 3 = -1 G(3) = F(2) + 2 * G(2) = 1 + 2 * 3 = 7

For n = 4: F(4) = 2 * F(3) - G(3) = 2 * (-1) - 7 = -9 G(4) = F(3) + 2 * G(3) = (-1) + 2 * 7 = 13

For n = 5: F(5) = 2 * F(4) - G(4) = 2 * (-9) - 13 = -31 G(5) = F(4) + 2 * G(4) = (-9) + 2 * 13 = 17

Now, we can calculate G(5) - F(5): G(5) - F(5) = 17 - (-31) = 17 + 31 = 48

Therefore, G(5) - F(5) is equal to 48.