Question

What are the values of these sums?

$\begin{array}{r}\sum _{k=1}^5 (k+1)\\ \sum _{j=0}^4 (-2{)}^j\\ \sum _{j=1}^{10} 3\\ \sum _{j=0}^8 \left(2^{j+1}-2^j\right)\end{array}$

Short answer

The values of the following sums are given below:

1. 20
2. 11
3. 30
4. 5 1 1

Step-by-step solution

Step 1:

Given: $\sum _{\mathrm{k}=1}^5 (\mathrm{k}+1)$

Compute the sum of this expression by substituting the values from the lower limit to the upper limit and then adding them together.

$\begin{array}{r}=(1+1)+(2+1)+(3+1)+(4+1)+(5+1)\\ =20\end{array}$

Step 2:

(b) Given : $\sum _{j=0}^4 (-2{)}^j$

Compute the sum of this expression by substituting the values from the lower limit to the upper limit and then adding them together.

$\begin{array}{r}=(-2{)}^0+(-2{)}^1+(-2{)}^2+(-2{)}^3+(-2{)}^4\\ =11\end{array}$

Step 3:

(c) Given : $\sum _{i=1}^{10} 3$

Compute the sum of this expression by substituting the values from the lower limit to the upper limit and then adding them together.

$\begin{array}{r}=3+3+3+3+3+3+3+3+3+3\\ =30\end{array}$

Step 4:

(d) Given: $\sum _{j=0}^8 \left(2^{j+1}-2^j\right)$

Compute the sum of this expression by substituting the values from the lower limit to the upper limit and then adding them together. Note that this is a telescoping sum.

$\begin{array}{r}=\left(2^1-2^0\right)+\left(2^2-2^1\right)+\left(2^3-2^2\right)+\left(2^4-2^3\right)+\left(2^5-2^4\right)+\left(2^6-2^5\right)+\left(2^7-2^6\right)+\left(2^8-2^7\right)+\left(2^9-2^8\right)\\ =2^1-2^0+2^2-2^1+2^3-2^2+2^4-2^3+2^5-2^4+2^6-2^5+2^7-2^6+2^8-2^7+2^9-2^8\end{array}$

Thus, by cancelling the like terms we get = 511