SOLUTION: 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.
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-> SOLUTION: 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.
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Question 1187141: 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. Answer by Boreal(15235) (Show Source):
You can put this solution on YOUR website! the first term is 9C0(1^9)(2x)^0=1
the second term is 9C1*(1^8)(2x)^1=18x
the third term is 9C2*(1^7)*(2x)^2=36*4x^2=144x^2
the fourth term is 9C3*(1^6)(2x)^3=84*8x^3=672x^3
here, x=0.01
so the sum is 1+0.18+0.0144+0.000672=1.195072
1.195092 is the actual answer to 6 decimal places
