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Let n be an odd natural number greater t...

Let `n` be an odd natural number greater than 1. Then , find the number of zeros at the end of the sum `99^n+1.`

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The correct Answer is:
2 zeroes
`1+99^(n) = 1+(100-1)^(n)`
`= 1+[.^(n)C_(0)100^(n) - .^(n)C_(1) 100^(n-1)+.^(n)C_(2) 100^(n+2)+"….."-.^(n)C_(n)]`
[`:'` n is odd]
`= 100[.^(n)C_(0) 100^(n-1)-.^(n)C_(1) 100^(n-2) + "….." + .^(n)C_(n-1)]`
`= 100 xx "integer"`
Hence, the number of zeros, at the end of the sum `99^(n) + 1` is `2`.
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