1i am curious to know your method because i solved it now and i am getting the ans as
[x]=2
x=\sum_{n=3}^{100}{\frac{1}{n^2 -4 }} \\ \color{red}then \ find \ [x]
my answer is coming out to be -3
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5 Answers
1
1i am getting 2.0sumthing
my method
x=\frac{1}{4}\sum_{n=3}^{100}{\frac{1}{n-2}+\frac{1}{n+2}} \\ now \\ \sum_{n=3}^{100}{\frac{1}{n-2}} = \int_{0}^{1}{1+x+x^2 +.............+x^{97}} \\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\int_{0}^{1}{\frac{x^{98}-1}{x-1}} \\ similarly \\ \sum_{n=3}^{100}{\frac{1}{n+2}} = \int_{0}^{1}{x^4 +x^5.............+x^{101}} \\ subtracting \ both \\ \frac{1}{4}\sum_{n=3}^{100}{\frac{1}{n-2}+\frac{1}{n+2}}=\int_{0}^{1}{(x^{98}-1)(x^2+1)(x+1).dx}
small edit : we are subtracting i wrote , but i have added :D,also please put a negative sign in front :)
1nice solution akari, u have used definite integration very cleverly
i also have another method :
1/(n2-4)=[1/(n-2)-1/(n-1)]-[1/(n+2)-1/(n+1)]+[1/(n-1)-1/n]-[1/(n+1)-1/n]
the given series is a telescopic series which we can sum to get the ans as - 2
1i dont think the ans can be negative as :
1/(n-2)-1/(n+2) will be positive
so the whole sum will be positive also
