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Question : 76 of 160
Marks:
+1,
-0
Solution:
Consider the integral,
P= [(n+1)(n+2)...(2n)] Apply log on both sides,
log‌P=[log()() ....()] =‌log(1+) =log(1+x) =[x‌log(1+x)]01−dx Further simplify the above,
log‌P=log‌2−(x−log(1+x))01 =log‌2−1+log‌2 =log‌4−1 =log() This implies,
P=
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