Let a, b and c be such that b(a+c) ≠ 0. If ∣∣
∣∣aa+1a−1−bb+1b−1cc−1c+1∣∣
∣∣+∣∣
∣
∣∣a+1b+1c−1a−1b−1c+1(−1)n+2a(−1)n−1b(−1)nc∣∣
∣
∣∣=0, then the value of n is
A
zero
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B
any even integer
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C
any odd integer
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D
any integer
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Solution
The correct option is C any odd integer ∣∣
∣∣aa+1a−1−bb+1b−1cc−1c+1∣∣
∣∣+(−1)n∣∣
∣∣a+1b+1c−1a−1b−1c+1a−bc∣∣
∣∣ =∣∣
∣∣aa+1a−1−bb+1b−1cc−1c+1∣∣
∣∣+(−1)n∣∣
∣∣a+1a−1ab+1b−1−bc−1c+1c∣∣
∣∣ =∣∣
∣∣aa+1a−1−bb+1b−1cc−1c+1∣∣
∣∣+(−1)n+2∣∣
∣∣aa+1a−1−bb+1b−1cc−1c+1∣∣
∣∣ This is equal to zero only if n + 2 is odd, i.e, n is an odd integer.