Using sigma notation. Find the partial sum?

Question

(46 is above the sigma sign, and n=1 is below) (9n+4)
Find the partial sum

a: 10,120
b: 9913
c. 455
d. 9729

Answer 1

if c is a constant…

∑ [f(n) + g(n)] = ∑ [f(n)] + ∑ [g(n)]

∑ [c*f(n)] = c*∑[f(n)]

(n=1 to m) ∑ [c] = cn

(n=1 to m) ∑ [n] = m(m + 1)/2



and so using these rules, we get:



(n=1 to 46) ∑ [9n + 4]

= (n=1 to 46) ∑ [9n] + (n=1 to 46) ∑ [4]

= 9*(n=1 to 46) ∑ [n] + (n=1 to 46) ∑ [4]

= 9*46(46 + 1)/2 + (n=1 to 46) ∑ [4]

= 9*46(47)/2 + (n=1 to 46) ∑ [4]

= 9*23(47) + (n=1 to 46) ∑ [4]

= 9*1081 + (n=1 to 46) ∑ [4]

= 9729 + (n=1 to 46) ∑ [4]

= 9729 + 4*46

= 9729 + 184

= 9913



so the answer is b

Answer 2

u can interpret the sum as the area under the line 9n+4 which forms a trapezoid.



so (9(1)+4+9(46)+4)46/2=9913



so the ans is b.

Answer 3

{1,46}E(9n+4)



split the sum into two sums

= {1,46}E(9n) + {1,46}E(4)



with the fist term (9*1+9*2+9*3+...+9*46); 9 can be factored out, so you are left with

9*(1+2+3+...+46)

the sum (not yet multiplying by 9) as you may remember from the story about Gauss (if not, ask a teacher because it is a fun story) can be computed as

(1/2)*(n^2 + n) where n=46

= 1081 (dont forget to multiply by 9)

1081*9 =9729



finally, one needs to add the second term {1,46}E(4), which is (4+4+4+...+4)

this of course is the same as 4*(46) =184



finally, 9729+180 = 9913 :D



answer: b: 9913