SOLUTION
Write out the expression
[tex]\begin{gathered} \sum ^3_{k\mathop=1}\lbrack1^k+(-1)^k\rbrack \\ \text{This implies k=1,2,3} \end{gathered}[/tex]
Then we substitute k=1 into the expression and obatin the value
[tex]\begin{gathered} k=1 \\ \lbrack1^1+(-1)^1\rbrack=\lbrack1+(-1)\rbrack=\lbrack1-1\rbrack=0 \end{gathered}[/tex]
Similarly,
[tex]\begin{gathered} k=2 \\ \lbrack1^2+(-1)^2\rbrack=\lbrack1+(1)\rbrack=\lbrack1+1\rbrack=2 \end{gathered}[/tex]
Then we also substitute the last value of k
[tex]\begin{gathered} k=3 \\ \lbrack1^3+(-1)^3\rbrack=\lbrack1+(-1)\rbrack=\lbrack1-1\rbrack=0 \end{gathered}[/tex]
finally, we take the sum of the result
[tex]\begin{gathered} 0+2+0=2 \\ \text{hence } \\ \sum ^3_{k\mathop{=}1}\lbrack1^k+(-1)^k\rbrack=2 \end{gathered}[/tex]
Therefore the summation of the expression from 1 to 3 for the values of k is 2
The right option is E (2).
