document.write( "Question 1125243: Use the binomial Theorem to determine the value of (1.03)^7 to four decimal places? Please help I don’t understand how to do it \n" ); document.write( "

Algebra.Com's Answer #741547 by Boreal(15235) $\"\"$ $\"About$

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(a+b)^7=a^7+7a^6b+21a^5b^2+35a^4b^3+35a^3b^4+21a^2b^5+7ab^6+b^7
\n" ); document.write( "(1.03)^7=(1+0.03)^7
\n" ); document.write( "1 to any power of course is 1
\n" ); document.write( "1+7*0.03+21*0.0009+35*0.000027+35*0.0000008+ and the rest will be 0, since 0.03^5 is about 2 x 10^(-8)\r
\n" ); document.write( "\n" ); document.write( "This is 1.21+0.0189+0.0009=1.2298
\n" ); document.write( "1.03^7 from calculator is 1.2299\r
\n" ); document.write( "\n" ); document.write( "The binomial expansion for (a+b)^n is a^nb^0 (the exponents have to add to n)
\n" ); document.write( "second term is nC1 (which is n) *a^(n-1)b^1
\n" ); document.write( "third term is nC2 (which is n/(n-2)!2!\r
\n" ); document.write( "\n" ); document.write( "the coefficients for exponent 7 are 1-7-21-35-35-21-7-1
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