$x=\frac{1}{16 \sqrt{17}+17 \sqrt{16}}+\frac{1}{17 \sqrt{18}+18 \sqrt{17}}+\ldots \ldots+\frac{1}{144 \sqrt{145}+145 \sqrt{144}}$ the value of $\mathrm{X}$ is approximately.
(A) 0.09
(B) 0.17
(C) 0.29
(D) 0.41
Please help me in the above. I tried but got to sqrt(a)-sqrt(b)/ab where a is the bigger of each term in the fraction but how to get the series to have a common denominator to cancel the terms.
Hint (re telescoping terms):
Try multiplying the first term by $$\frac{17\sqrt{16} - 16\sqrt{17}}{17\sqrt{16} - 16\sqrt{17}}$$ and see what happens.