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1/1+root2+1/rrot2+root3+1/root3+root4+1/root4+root5+1/root5+root6+1/root6+root7+1/root7 + root8+1/root8+root9 = 2 . prove that

please help me



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Notice that

The given sum is same as
(\sqrt{2}-\sqrt{1})+(\sqrt{3}-\sqrt{2})+\cdots+(\sqrt{8}-\sqrt{7})+(\sqrt{9}-\sqrt{8})\\\text{cancel the~middle terms and get}\\=-\sqrt{1}+\sqrt{9}\\=-1+3\\=2
2 5 2
second step
good, so did you get how 1/(sqrt(n+1)+sqrt(n)) becomes sqrt(n+1) - sqrt(n) ?
yes i understood the whole sum thanks
sounds great! it seems the latex is not fully functional here, im not able to put the "cancel" bars to show exactly how all the middle terms cancel out
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