Simplify: n+(n+1)+(n+2)+...+[n+(n-1)]+...+(n+2)+(n+1)+n
2 answers:
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All exercises involve the same concept, so I'll show you how to do the first, then you can apply the exact same logic to all the others.
The first thing you need to know is that, when a certain quantity multiplies a parenthesis, you can distribute that number to every element in the parenthesis. This means that
So, is multiplying the parenthesis involving
and
, and we distributed it:
multiplies both
and
in the final result.
Secondly, you have to know how to recognize like terms, because they are the only terms you can sum. Two terms can be summed if they have the same literal expression. So, for example, you cannot sum , and neither
exponents count.
But you can su, for example,
or
So, take for example exercise 9:
We distribute the 1.2 through the first parenthesis:
And you can distribute the negative sign through the second parenthesis (it counts as a -1 to distribute):
So, the expression becomes
Now sum like terms:
Answer:
A
Step-by-step explanation:
Glad to help
If you will have any questions about the way I solved it,
don't hesitate to ask
