Question

Help Please!!!

Expand the series and evaluate:

PROBLEM BELOW !

Answer 1

\[\sum_{k=1}^{4} (-1) ^{k+1}( \frac{12 }{ k ^{2} })\]

Answer 2

just input k = 1 till k =4
for k = 1, -> (-1)^(1+1)*(12/1^2)= (-1)^2 * 12/1 = 1 * 12 = 12
for k = 2, -> (-1)^(2+1)*(12/2^2)= (-1)^3 * 12/4 = -1 * 3 = -3
for k = 3, -> (-1)^(3+1)*(12/3^2)= (-1)^4 * 12/9 = 1 * 12/9 = 4/3
for k = 4, -> (-1)^(4+1)*(12/4^2)= (-1)^5 * 12/16= -1 * 12/16 =- 3/4

now, can you evaluate the value of
12 - 3 + 4/3 - 3/4 = .... ?

Answer 3

\[9\frac{ 7 }{ 12 } thanks\]

Answer 4

that;s right :)