11 summation n=1(2+3^n) evaluate?

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11 summation n=1(2+3^n) evaluate?

2007-02-22 14:04:02 · 2 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics

2 answers

Sum ( n= 1 to 11) (2+3^n)

= Sum ( n= 1 to 11) (2) +Sum ( n= 1 to 11) (3^n)

= 2*11 + ( 3 + 3^2 + .....+ 3^11)

= 22 + 3(1 + 3 + 3^2 + ....+ 3^10)

= 22 + 3*(3^11 -1)/(3-1)

= 22 + 3*177146/2

= 22 + 265719

= 265741

2007-02-23 02:43:04 · answer #1 · answered by K Sengupta 4 · 0⤊ 0⤋

May U mean : summation from n=1 to n=11 [(2+3^n)]
= sum from n=1 to 11 [2] + sum from n=1 to 11 [3^n]
= 11Ã2 + [ 3(3^11 -1)/(3-1)] (geometric sequence )
= 22 + 265719 = 265741

2007-02-22 21:14:59 · answer #2 · answered by a_ebnlhaitham 6 · 0⤊ 0⤋