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निम्नलिखित अनन्त श्रेणी का योग ज्ञात कीजिए | `Sigma_(n=0)^(oo)(1)/(n!)[Sigma_(k=0)^(n){(k+1)int_(0)^(1)2^(-(k+1)x)dx}]`

Answer» `int_(0)^(1)2^(-(k+1)x).dx=[(2^(-(k+1)x))/(-(k+1)log2)]_(0)^(1)=(2^(-(k+1)))/(-(k+1)log2)`
`:. underset(k=0)overset(n)Sigma(k+1){(2^(-(k+1))-1)/(-(k+1)log2)}=(1)/(log2)underset(k=0)overset(n)Sigma{1-(1)/(2^(k+1))}`
`=(1)/(log2)[(n+1)-{(1)/(2)+(1)/(2^(2))+(1)/(2^(3))+...+(n+1)पद}]`
`=(1)/(log2)[(n+1)-((1)/(2){1-((1)/(2))^(n+1)})/((1-(1)/(2)))]`
`=(1)/(log2)[(n+1)-(1-2^(-(n+1)))]=(1)/(log2)[n+((1)/(2))^(n+1)]`
`:. " "S=(1)/(log2)underset(n=0)overset(oo)Sigma[(n)/(n!)+(1)/(2).(((1)/(2))^(n))/(n!)]=(1)/(log2)underset(n=0)overset(oo)Sigma[(1)/((n-1)!)+(1)/(2).(x^(n))/(n!)]`
जहाँ`x=(1)/(2)`
`=(1)/(log2)[e+(e^(x))/(2)]=(1)/(log2)[e+(1)/(2)sqrt(e)]`


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