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This is the problem: Write down the binomial expansion of $(1+k/100)^{1/2}$ in ascending powers of $k$, up to and including the $k^3$ term.

Use the value $k=8$ to find an approximation to five decimal places for $3^{1/2}$.

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    $\begingroup$ where are you stuck? $\endgroup$ – Masacroso Feb 2 '17 at 9:11
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We have that $$(x+y)^r = \sum_{k=0}^{\infty} \binom {r}{k} x^{r-k}y^{k} = x+ rx^{r-1}y + \frac {r (r-1)}{2!}x^{r-2}y^2 +\cdots $$ This is the Newton's Generalised binomial theorem. Now put $x=1$ and $y=\frac {k}{100} $ and put $r =\frac {1}{2} $ and see what you get. Hope it helps.

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