When P is a natural number, then Pn+1+(P+1)2n−1 is divisible by
A
P
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B
P2+P
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C
P2+P+1
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D
P2+1
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Solution
The correct option is CP2+P+1 For n = 2, we get, Pn+1+(P+1)2n−1=P3+(P+1)3 =P3+P3+1+3P2+3P=2P3+3P2+3P+1 Which is divisible by P2+P+1 Given result is true for all nϵN.