Answer:
6
Step-by-step explanation:
We need to evaluate :-
[tex]\rm\implies \displaystyle\rm\sum^4_n (-1)^n (3n + 2 ) [/tex]
Here the [tex]\Sigma[/tex] is the sum operator . And here we need to find the sum from n = 1 to n = 4 . We can write it as ,
[tex]\rm\implies (-1)^1 ( 3*1 +2) + (-1)^2 ( 3*2+2) + (-1)^3(3*3+2) + (-1)^4(3*4+2) [/tex]
Now we know that for odd powers of -1 , we get -1 and for even powers we get 1 . Therefore ,
[tex]\rm\implies -1 ( 3 + 2 ) + 1 (6+2)+-1(9+2)+1(12+2)[/tex]
Now add the terms inside the brackets and then multiply it with the number outside the bracket . We will get ,
[tex]\rm\implies -1 * 5 + 1 * 8 + -1*11 + 1*14 \\\\\rm\implies -5 + 8 - 11 + 14 \\\\\rm\implies\boxed{\quad 6 \quad}[/tex]
Hence the required answer is 6.
