Question

Find the value of each of these sums.

$\begin{array}{r}\left(\mathrm{a}\right)\sum _{\mathrm{j}=0}^8 \left(1+(-1{)}^{\mathrm{j}}\right)\\ \left(\mathrm{b}\right)\sum _{\mathrm{j}=0}^8 \left(3^{\mathrm{j}}-2^{\mathrm{j}}\right)\\ \left(\mathrm{c}\right)\sum _{\mathrm{j}=0}^8 \left(2\cdot 3^{\mathrm{j}}+3\cdot 2^{\mathrm{j}}\right)\\ \left(\mathrm{d}\right)\sum _{\mathrm{j}=0}^8 \left(2^{\mathrm{j}+1}-2^{\mathrm{j}}\right)\end{array}$

Short answer

The values of the following sums are given below:

1. 10
2. 9330
3. 21215
4. 511

Step-by-step solution

Step 1: 

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

Note that since summation is a linear operator, then distributive property can be used. Use the geometric series sum formulas twice to compute the sum.

Note that $a_0=(-1)°$= 1 and n = 8for the first sum.

Also,$a_0=(-1{)}^0=1,n=8\text{and}r=-1$

$\begin{array}{r}\sum _{j=0}^8 \left(1+(-1{)}^j\right)\\ =\sum _{j=0}^8 \left(1\right)+\sum _{j=0}^8 (-1{)}^j\\ =a_0(n+1)+a_0\left[\frac{r^{n+1}-1}{r-1}\right]\\ =1(8+1)+1\left[\frac{(-1{)}^9-1}{-1-1}\right]\\ =9+1\\ =10\end{array}$

Step 2:

$\text{Given:}\sum _{j=0}^8 \left(3^j-2^j\right)$

Note that since summation is a linear operator, then distributive property can be used. Use the geometric series sum formulas twice to compute the sum.

Note that ${\mathrm{a}}_0=3^0=1,\mathrm{r}=3\text{and}\mathrm{n}=8$for the first sum.

Also, ${\mathrm{a}}_0=2^0=1,\mathrm{r}=2\text{and}\mathrm{n}=8$for the second sum.

$\begin{array}{r}\sum _{j=0}^8 \left(3^j-2^j\right)\\ =\sum _{j=0}^8 \left(3^j\right)-\sum _{j=0}^8 2^j\\ =a_0\left[\frac{r^{n+1}-1}{r-1}\right]-a_0\left[\frac{r^{n+1}-1}{r-1}\right],\\ =1\left[\frac{3^9-1}{3-1}\right]-1\left[\frac{2^9-1}{2-1}\right]\\ =9841-511\\ =9330\end{array}$

Step 3:

$\text{Given:}\sum _{\mathrm{j}=0}^8 \left(2\cdot 3^{\mathrm{j}}+3\cdot 2^{\mathrm{j}}\right)$

Note that since summation is a linear operator, then distributive property can be used. Use the geometric series sum formulas twice to compute the sum.

Note that $a_0=2\left(3^0\right)=2,r=3\text{and}n=8$for the first sum.

Also, $a_0=3\left(2^0\right)=3,r=2,\text{and}n=8$for the second sum.

$\begin{array}{r}\sum _{j=0}^8 \left(2\cdot 3^j+3\cdot 2^j\right)\\ =2\sum _{j=0}^8 \left(3^j\right)+3\sum _{j=0}^8 \left(2^j\right)\\ =a_0\left[\frac{r^{n+1}-1}{r-1}\right]+a_0\left[\frac{r^{n+1}-1}{r-1}\right]\end{array}$

$\begin{array}{r}=2\left[\frac{3^9-1}{3-1}\right]+3\left[\frac{2^9-1}{2-1}\right]\\ =2\left(9841\right)+3\left(511\right)\\ =21215\end{array}$

Step 4:

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

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