document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1187141>Question 1187141</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Obtain the first 4 terms in the expansion of (1+2x)^9 in ascending powers of x.Use this expansion to find an approximate value of (1.02)^9. </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #819233 by </B><A HREF=/tutors/aboutme.mpl?userid=Boreal><B>Boreal(15235)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=Boreal><IMG SRC=http://www.algebra.com/images/aboutme-small.gif BORDER=0 alt=\"About Me\" VALIGN=MIDDLE></A>&nbsp;<IMG SRC=http://www.algebra.com/cgi-bin/show-face.mpl?user=Boreal><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=819233\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> the first term is 9C0(1^9)(2x)^0=1<BR>\n" );
document.write( "the second term is 9C1*(1^8)(2x)^1=18x<BR>\n" );
document.write( "the third term is 9C2*(1^7)*(2x)^2=36*4x^2=144x^2<BR>\n" );
document.write( "the fourth term is 9C3*(1^6)(2x)^3=84*8x^3=672x^3\r<BR>\n" );
document.write( "\n" );
document.write( "here, x=0.01<BR>\n" );
document.write( "so the sum is 1+0.18+0.0144+0.000672=1.195072<BR>\n" );
document.write( "1.195092 is the actual answer to 6 decimal places </SPAN>\n" );
document.write( "</TD></TR></TABLE>\n" );