Question

done ty

Answer 1

how do I solve sigma notations? I need to do more like this

Answer 2

You should simply do substitution,
\[\sum_{n=3}^{12} (-5n-1)=-\sum_{n=3}^{12} (5n+1)=(5\cdot 3+1)+(5\cdot 4+1)+\ldots+5\cdot 12+1=385\]

Answer 3

\[=385\]

Answer 4

Why do we change the -5n-1 to plus though?

Answer 5

with the minus sign, obviusly. sum=-385

Answer 6

It is not necessary to change the sign. As both numbers has the minus, I extract common factor the sign, and put it out the sum. But you must take it into account later.

Answer 7

You can also do the following reasoning,
\[-\sum_{n=3}^{12}(5n+1)=-(5\sum_{n=3}^{12}n+\sum_{n=3}^{12}1)=\\
=-(5\cdot75+10)=-385\]

Answer 8

so n goes from 1 to 14 in this one I would add 1+2+3+4+5+6+7+8+9+10+12+11+13+ 14 to get the sum rigtht?

Answer 9

Yes, but take into account the 3 and the 2,
\[\sum_{n=1}^{14}(3n+2)=(3+2)+(3\cdot 2+2)+\ldots+(3\cdot14+2)\]

Answer 10

13+...+ 44 = 57 do I add it with 105 or subtract

Answer 11

The same "trick" as before is avaliable here,
\[\sum_{n=1}^{14}(3n+2)=3\sum_{n=1}^{14}n+\sum_{n=1}^{14}2=3\cdot105+28=343\]

Answer 12

Oh okay I think I got it now I will go keep practicing thank you very much though for the help.

Answer 13

You're welcome.