Question

Please help I’ll give brainliest

Answer 1

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