Q.

Let n be a positive integerA=∑k=0n−1kCk   n12k+34k+78k+1516k+3132k and if 63A=1−1230 then the value of n is

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answer is 6.

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Detailed Solution

A=∑k=0n−1kCk   n12k+34k+78k+1516k+3132k=12n+14n+18n+116n+132n=12n+122n+123n+124n+125n=12n125n−112n−1  =125n25n−12n−1 63A=6325n25n−12n−1=632n−11−125n = 1−1230 (given)⇒n=6
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