InterviewSolution
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InterviewSolution
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Let `I_(n)=int_(0)^(pi//2)x^(n)cosxdx,` where in is a non-negative integer Then `sum_(n=2)^(oo)((I_(n))/(n!)+(I_(n-2))/((n-2)!))` equals-A. `e^(pi//2)-1-(pi)/(2)`B. `e^(pi//2)-1`C. `e^(pi//2)-(pi)/(2)`D. `e^(pi//2)` |
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Answer» Correct Answer - A `I_(n)=int_(0)^(pi//2)x_(1)^(n)underset(II)cosxdx` `=x^(n)sinx|_(0)^(pi//2)-int_(0)^(pi//2)nx^(n-1)sinxdx` `=((pi)/(2))^(2)-0-(nx^(n-1)(-cosx)|_(0)^(pi//2)-int_(0)^(pi//2)n(n-1)x^(n-2)(-cosx)dx` `=((pi)/(2))^(n)-0-(n-1)int_(0)^(pi//2)x^(n-2)cosxdx` `I_(n)=((pi)/(2))^(n)-n(n-1)I_(n-2)` `underset(n=2)overset(oo)sum((I_(n))/(n!)+(I_(n-2))/((n-2)!))=underset(n=2)overset(oo)sum((((pi)/(2))^(2)-n(n-1)I_(n-2))/(n!)+(I_(n-2))/((n-2)!))` `underset(n=2)overset(oo)sum(((pi)/(2))^(n)(1)/(n!))` `=((pi)/(2))^(2)(1)/(2!)+((pi)/(2))^(3)(1)/(3!)+((pi)/(2))^(4)(1)/(4!)+....` `e^(pi//2)-1-((pi)/(2))` |
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