my friend and I have no clue to solve the mentioned strategy
a and b must have to be equal to 1 because of the definition of fibonacci. And we is thinking it has to do something about of: $X_{n}= r_{1}+r_{2}$
my friend and I have no clue to solve the mentioned strategy
a and b must have to be equal to 1 because of the definition of fibonacci. And we is thinking it has to do something about of: $X_{n}= r_{1}+r_{2}$
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d) If the consequences become are (1), then they get the equation below. $$ r ^ {n-1} = b_1 \cdot r ^ {n-2} + ... + b_m \cdot r ^ {n- (m + 1)} $$ Since the consequences result (1), then it must also apply to $ x_n = r_1 + ... + r_m $ moeter (1). If the consequences become (1), then they must also require (1) if multiplied and constantly on, as shown in the equation below. $$ c_1 \cdot r ^ {n-1} = c_1 \cdot (b_1 \cdot r ^ {n-2} + ... + b_m \cdot r ^ {n- (m + 1)}) = c_1 \cdot b_1 \cdot r ^ {n-2} + ... + c_1 \cdot b_m \ cdot r ^ {n- (m + 1)} $$ Since the consequences supplement (1) with a constant multiplied by, it must also apply that $ x_n = c_1 \cdot r_1 + .... + c_m \cdot r_m $ (1).
any tips to help argue for this question