Evaluate :
(i) `sum_(n=1)^(10)(2+3^(n))" "`(ii) `sum_(k=1)^(n)[2^(k)+3^((k-1))]" "`(iii) `sum_(n=1)^(8) 5^(n)`

Answer

(i) 29544 (ii) `1/2[2^((n+2))+3^(n)-5]` (iii) 488280

Explanation

(i) `sum_(n=1)^(10) (2+3^(n))=(2+3^(1))+(2+3^(3))+(2+3^(3))+...+(2+3^(10))`
`=(2+2+2+..."taken 10 times")+(3+3^(2)+3^(3)+...+3^(10))`
`=(2xx10)+{3xx((3^(10)-1)/(3-1))}`
`=20+((59049-1)/2)=(20+29524)=29544`.
(ii) `sum_(k=1)^(n)[2^(k)+3^((k-1))]=(2^(1)+3^(0))+(2^(2)+3^(1))+(2^(3)+3^(2))+`... up to n terms
`=(2^(1)+2^(2)+2^(3)+..."to n terms")+(1+3+3^(2)+..."to n terms")`
`=2((2^(n)-1)/(2-1))+1. ((3^(n)-1)/(3-1))`
`=2(2^(n)-1)+1/2(3^(n)-1)=2^((n+1))-2+1/2 . 3^(n)-1/2`
`=2^((n+1))+1/2. 3^(n)-5/2=2^((n+1))+1/2 (3^(n)-5)=1/2 [2^((n+2))+3^(n)-5]`.
(iii) `sum_(n=1)^(8) 5^(n)=[5+5^(2)+5^(3)+..."to 8 terms"]`
`=5((5^(8)-1)/(5-1))=(5(390625-1))/4=(5xx390624)/4`
`=(5xx97656)=488280`.