Hello from MrBillDoesMath!
Answer: The second bullet point from the top.
(43/2) * (133/8 + 7/8)
Discussion:
The 43rd partial sum =
(3/8)* ( sum from n = 1 to 43 of n) + 43 * (1/2)
As the sum of the first "n" integers is n*(n+1)/2, the first sum above, with n = 43, becomes
(3/8)* ( 43 * (43+1)/2) + 43 * (1/2)
Factor 43 from each term
43 * ( (3/8) * (43+1)/2) + (1/2) )
Factor 1/2 from each term
(43/2) * ( (3/8)* 44 + 1 ) =>
(43/2) * ( (3 * 44)/8 + 1 ) =>
(43/2) * ( 132/8 + 1 )
Substituting 132 = 133 -1 above
(43/2) * ( (133-1)/8 + 1 ) =>
(43/2) * ( 133/8 - (1/8) + 1) =>
(43/2) * (133/8 + 7/8)
which is the second bullet point.
Thank you,
MrB
