document.write( "Question 1091387: please help me on the following qn:\r
\n" ); document.write( "\n" ); document.write( " Obtain the first four terms of the expansion of (1+8x)^0.5 in ascending powers of x by putting x=0.01, obtain the value of (3)^0.5 correct to five decimal places \n" ); document.write( "

Algebra.Com's Answer #706721 by Fombitz(32388) $\"\"$ $\"About$

You can put this solution on YOUR website!
$\"%281%2B8x%29%5E%280.5%29=1%2B4x-8x%5E2%2B32x%5E3\"$
\n" ); document.write( "So then substituting,
\n" ); document.write( " $\"%281%2B8%280.01%29%29%5E%280.5%29=1%2B4%280.01%29-8%280.01%29%5E2%2B32%280.01%29%5E3\"$
\n" ); document.write( " $\"sqrt%281.08%29=1%2B%280.04%29-%280.0008%29%2B%280.000032%29\"$
\n" ); document.write( " $\"sqrt%281.08%29=1.039232%29\"$
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\n" ); document.write( " $\"1%2B8x=3\"$
\n" ); document.write( " $\"8x=2\"$
\n" ); document.write( " $\"x=1%2F4\"$
\n" ); document.write( "However, the Taylor series expansion at $\"x=0\"$ only converges when $\"abs%288x%29%3C1\"$
\n" ); document.write( " $\"abs%28x%29%3C1%2F8\"$ \r
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