This summation will repeat three times, for $\displaystyle n=4,5,6$.

Substitute the first term into the parentheses, repeat, and sum the process for the next three terms.

$\displaystyle \sum_{n=3}^{6} (1-6n)$

$\displaystyle =(1-6(3))+(1-6(4))+(1-6(5))+(1-6(6))$

Simplify each term.

$\displaystyle (-17)+(-23)+(-29)+(-35) =-104$

The answer is:  $\displaystyle -104$

Expand the summation sign.  Start with zero as the first index, then one, and finally two.  Since two is the top digit, the summation will stop.

$\displaystyle \sum_{0}^{2}2n^6=2(0)^6+2(1)^6+2(2)^6$

Simplify these terms.

$\displaystyle 0+2+2(64) = 2+128 = 130$

The answer is:  $\displaystyle 130$

This notation is asking us to add $\displaystyle k^2 - 7$ for all integer values of k between 1 and 4: 

$\displaystyle (1^2 - 7 ) + (2^2 - 7 ) + (3^2 - 7 ) + (4^2 - 7 )$

$\displaystyle (1-7) + (4-7) + (9-7) + (16-7 ) = -6 + -3 + 2 + 9 = 2$

This expression is asking us to add the expressions $\displaystyle \binom{4}{n}$ for every integer value of n from 0 to 4:

$\displaystyle \binom{4}{0 } + \binom{4}{1} + \binom{4}{2} + \binom{4}{3} + \binom{4}{4}$

$\displaystyle \frac{4!}{0! \cdot 4!} + \frac{4!}{1!\cdot3! } + \frac{4!}{2! \cdot 2! } + \frac{4! } { 3! \cdot 1! } + \frac{4!}{4! \cdot 0!}$

$\displaystyle 1 + 4 + 6 + 4 + 1 = 16$

$\displaystyle \sum_{k=1}^{3} 20 \cdot (\frac{1}{2} ) ^k$means that we are adding half of 20, plus half of that, plus half of that.

1: $\displaystyle 20 \cdot (\frac{1}{2} ) = 10$

2: $\displaystyle 10(\frac{1}{2} ) = 5$

3: $\displaystyle 5(\frac{1}{2} ) = 2.5$

$\displaystyle 10 + 5 + 2.5 = 17.5$

This is asking us to add $\displaystyle n-5$ for n = 3 and 4.

$\displaystyle (3 -5 ) + (4 - 5 ) = -2 + -1 = -3$

This is asking us to plug in the integers between 0 and 5, then add these numbers together.

$\displaystyle (\frac{1}{2} (0)^2 +1 ) + (\frac{1}{2} (1)^2 +1 ) + (\frac{1}{2} (2)^2 +1 ) + (\frac{1}{2} (3)^2 +1 ) + (\frac{1}{2} (4)^2 +1 ) +(\frac{1}{2} (5)^2 +1 )$

$\displaystyle = (0+1) + (.5+1) + (2+1) + (4.5+1) + (8+1) + (12.5+1) = 33.5$

This is asking us to add 6 plus two thirds of 6, plus two thirds of that, etc.

$\displaystyle 6 + 6 \cdot \frac{2}{3} + 6 \cdot\left ( \frac{2}{3} \right )^2 + 6 \cdot \left (\frac{2}{3} \right ) ^3 + 6 \cdot \left (\frac{2}{3} \right )^4$

$\displaystyle = 6 + 4 + 2.\overline{6} + 1. \overline{7} + 1.\overline{185} = 15. \overline{629}$

This is asking us to substitute integer values between 4 and 8 for n, and then add the results.

$\displaystyle (4+1)^2 + (5+1)^2 + (6+1)^2 + (7+1)^2 + (8+1)^2$

$\displaystyle =25 + 36+49+64+81 = 255$

First, evaluate the sigma expression. It is asking us to plug in for k all of the integers between 0 and 4:

$\displaystyle \left [2(0)-1 \right ] + \left [2(1)-1 \right ] + \left [2(2) - 1 \right ]+ \left [2(3) - 1 \right ] + \left [2(4) -1 \right ]$

$\displaystyle = -1 + 1 + 3 + 5 + 7 = 15$

Now, multiply by $\displaystyle \frac{1}{3} : \frac{1}{3} \cdot 15 = 5$