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Question : 76 of 160
Marks:
+1,
-0
Solution:
L=[]1∕n=(∏K=1n(1+))1∕nL=e(ln(1+))...Sum can be approximated by an integral in the limit as
n⟶∞⇒ln(1+)≈ln(1+x2)dxUsing integration by parts
ln(1+x2)dx=[ln(1+x2)⋅x]01−∫dx=ln2−2dx+2dx=ln2−2+So, from Eq. (i), we get
L=eln2−2+=2−2=2()
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