How do you use summation formulas to rewrite #\sum _ { i = 1} ^ { n } \frac { 8i + 9} { n ^ { 2} }#?

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Answer 1 · 2017-06-12T02:26:55

#sum_(i=1)^n (8i+9)/n^2#

#1/n^2 sum_(i=1)^n 8i+9#

#1/n^2[ sum_(i=1)^n 8i +sum_(i=1)^n 9]#

#1/n^2[ 8 sum_(i=1)^n i +sum_(i=1)^n 9]#

#sum_(i=1)^n i=((n)(n+1))/2#, since it is an arithmetic series with difference of 1 and first term of 1.

#sum_(i=1)^n 9=9n#, this is the summation of a constant where
#c=9#.

Putting it all together:

#1/n^2[ (8(n)(n+1))/2 +9n]=(4n+4)/(n)+9/n=((4n+4)+(9))/(n)=(4n+13)/(n)#