Question 1198089
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{{{sum(log((5r)),r=1,n) = log((5*1))+log((5*2))+log((5*3))+matrix(1,1,"...")+log((5*n))}}}


{{{sum(log((5r)),r=1,n=20) = log((5*1))+log((5*2))+log((5*3))+matrix(1,1,"...")+log((5*20))}}}


{{{sum(log((5r)),r=1,n=20) = (log((5))+log((1)))+(log((5))+log((2)))+(log((5))+log((3)))+matrix(1,1,"...")+(log((5))+log((20)))}}} Use the rule that log(A*B) = log(A)+log(B)


{{{sum(log((5r)),r=1,n=20) = (log((5))+log((5))+log((5))+matrix(1,1,"...")+log((5)))+(log((1))+log((2))+log((3))+matrix(1,1,"...")+log((20)))}}} See note1 below


{{{sum(log((5r)),r=1,n=20) = 20*log((5))+(log((1))+log((2))+log((3))+matrix(1,1,"...")+log((20)))}}} 


{{{sum(log((5r)),r=1,n=20) = 20*log((5))+log((matrix(1,1,"1*2*3*...*20")))}}} See note2 below


{{{sum(log((5r)),r=1,n=20) = 20*log((5))+log((matrix(1,1,"20!")))}}}


{{{sum(log((5r)),r=1,n=20) = 32.365524703598}}} which is approximate


It seems strange that your teacher is getting {{{S[20] = 146.78}}}, so I have a feeling I might be mis-interpreting the given expression.


Footnotes:<ul><li>note1: The first parenthesis grouping has 20 copies of "log(5)" added together, which leads to 20*log(5) in the next line.</li><li>note2: Use the rule log(A)+log(B) = log(A*B). The log(1*2*3*...*20) leads to log(20!) where the exclamation mark indicates factorial.</li></ul>
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