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Question
If A=(1011] and I=(1001], then which of the following holds for all n∈N?
If A=(1011] and I=(1001], then which of the following holds for all n∈N?
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Solution
The correct option is A An=nA−(n−1)I
We shall prove by principle of mathematical induction that first option is correct.Let P(n) be the statement,P(n): An=nA−(n−1)IConsider P(1)LHS=A and RHS=A−(1−1)I=AThus P(n) is true for n=1.Assume that P(k) is true for some k∈N.Then Ak=kA−(k−1)I ...(1)Now, we will show that P(k+1) is true whenever P(k) is true.Consider Ak+1=Ak.A =(kA−(k−1)I).A (Using (1)] =kA2−(k−1)AA2=A.A = (1011](1011]=(1021]we get A2=2A−ISo, Ak+1=k(2A−I)−(k−1)A =2kA−kI−(k−1)A =(2k−k+1)A−kI =(k+1)A−((k+1)−1)IThus, P(k+1) is true whenever P(k) is true.Hence, by principle of mathematical induction,P(n) is true for all natural number.Hence, the first option is correct.Then, clearly third option is wrong.Now, consider the second option.An=2n−1A−(n−1)IFor n=2, LHS=A2 =(1021] and RHS=2A−I=(1021] ⇒P(n) is true for n=2.
For n=3, LHS=A3=(1031]and RHS=4A−2I=(2042]Thus, for n=3, LHS≠RHS
Similarly, we can show that the fourthoption is not true for n=2.
We shall prove by principle of mathematical induction that first option is correct.Let P(n) be the statement,P(n): An=nA−(n−1)IConsider P(1)LHS=A and RHS=A−(1−1)I=AThus P(n) is true for n=1.Assume that P(k) is true for some k∈N.Then Ak=kA−(k−1)I ...(1)Now, we will show that P(k+1) is true whenever P(k) is true.Consider Ak+1=Ak.A =(kA−(k−1)I).A (Using (1)] =kA2−(k−1)AA2=A.A = (1011](1011]=(1021]we get A2=2A−ISo, Ak+1=k(2A−I)−(k−1)A =2kA−kI−(k−1)A =(2k−k+1)A−kI =(k+1)A−((k+1)−1)IThus, P(k+1) is true whenever P(k) is true.Hence, by principle of mathematical induction,P(n) is true for all natural number.Hence, the first option is correct.Then, clearly third option is wrong.Now, consider the second option.An=2n−1A−(n−1)IFor n=2, LHS=A2 =(1021] and RHS=2A−I=(1021] ⇒P(n) is true for n=2.
For n=3, LHS=A3=(1031]and RHS=4A−2I=(2042]Thus, for n=3, LHS≠RHS
Similarly, we can show that the fourthoption is not true for n=2.