求1/(√2+1)+1/(√3+√2)+…+1/(√n+√n-1)的值?

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求1/(√2+1)+1/(√3+√2)+…+1/(√n+√n-1)的值?

求1/(√2+1)+1/(√3+√2)+…+1/(√n+√n-1)的值?
求1/(√2+1)+1/(√3+√2)+…+1/(√n+√n-1)的值?

求1/(√2+1)+1/(√3+√2)+…+1/(√n+√n-1)的值?
因为1/(√n+√n-1)
=(√n-√n-1)/[(√n+√n-1)*(√n-√n-1)]
=√n-√(n-1)
所以1/(√2+1)+1/(√3+√2)+…+1/(√n+√n-1)
=(√2-1)+(√3-√2)+(√4-√3)+.+[√n-√(n-1)]
=√2-1+√3-√2+√4-√3+.+√n-√(n-1)
=√n-1