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Question

The value of limn{(n3+1)(n3+23)(n3+33)........(n3+n3)n3}1n is equal to α.eβx31+x3dx, where αϵN,βϵR,then find αβ.___


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Solution

y=limn{(n3+1)(n3+23)(n3+33)........(n3+n3)n3}1n
Taking log on both sides, we get
logy=limn1n{nr=1log(n3+r3)logn3}logy=limn1nnr=1log[1+(rn)3]logy=10log(1+x3)dxlogy=[log(1+x3).x]10103x21+x3xdxlogy=log2310x21+x3xdx y=2.e310x31+x3αβ=2(3)=5


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