If A = {$$6^{2n} -35n - 1$$}, where $$n$$ = 1,2,3,... and B = {35($$n$$-1)}, where $$n$$ = 1,2,3,... then whic

Question

If A = {$$6^{2n} -35n - 1$$}, where $$n$$ = 1,2,3,... and B = {35($$n$$-1)}, where $$n$$ = 1,2,3,... then which of the following is true?

Accepted Answer

Every member of A is in B and at least one member of B is not in A. If we carefully observe set A, then we find that 6 2 n − 35 n − 1 6^{2n} -35n - 1 6 2 n − 35 n − 1 is divisible by 35. So, set A contains multiples of 35. However, not all the multiples of 35 are there in set A, for different values of n n n . For n = 1 n = 1 n = 1 , the value is 0, for n = 2 n = 2

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If A = { $6^{2n} -35n - 1$ }, where $n$ = 1,2,3,... and B = {35( $n$ -1)}, where $n$ = 1,2,3,... then which of the following is true?

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If A = {62n−35n−16^{2n} -35n - 162n−35n−1}, where nnn = 1,2,3,... and B = {35(nnn-1)}, where nnn = 1,2,3,... then which of the following is true?

If A = { $6^{2n} -35n - 1$ }, where $n$ = 1,2,3,... and B = {35( $n$ -1)}, where $n$ = 1,2,3,... then which of the following is true?

If A = { $6^{2n} -35n - 1$ }, where $n$ = 1,2,3,... and B = {35( $n$ -1)}, where $n$ = 1,2,3,... then which of the following is true?