Chapter 1.1, Problem 1E

State whether the given sums are equal or unequal.

1. ${\displaystyle \underset{i=1}{\overset{10}{\sum }}i}$ and ${\displaystyle \underset{k=1}{\overset{10}{\sum }}k}$

${\displaystyle \underset{i=1}{\overset{10}{\sum }}i}$ and ${\displaystyle \underset{i=6}{\overset{15}{\sum }}\left(i-5\right)}$

${\displaystyle \underset{i=1}{\overset{10}{\sum }}i\left(i-1\right)}$ and ${\displaystyle \underset{j=0}{\overset{9}{\sum }}\left(j+1\right)j}$

${\displaystyle \underset{i=1}{\overset{10}{\sum }}i\left(i-1\right)}$ and ${\displaystyle \underset{k=1}{\overset{10}{\sum }}\left(k^2-k\right)}$

To determine

To check: whether the given sums ${\displaystyle \sum _{i=1}^{10}i}$ and ${\displaystyle \sum _{k=1}^{10}k}$ are equal or unequal.

Answer to Problem 1E

Both the sums ${\displaystyle \sum _{i=1}^{10}i}$ and ${\displaystyle \sum _{k=1}^{10}k}$ are equal.

Explanation of Solution

Given information:

The given sums are ${\displaystyle \sum _{i=1}^{10}i}$ and ${\displaystyle \sum _{k=1}^{10}k}$ .

Concept used:

Summation expansion formula:

  ${\displaystyle \sum _{i=1}^n{a}_i=a_1+}{a}_2+a_3+a_4+\mathrm{.........}+a_n$

Calculation:

We will expand both sums.

  ${\displaystyle \sum _{i=1}^{10}i=1+2+3+4+5+6+7+8+9+10}$

Now, we can add numbers.

  ${\displaystyle \sum _{i=1}^{10}i=55}$

  ${\displaystyle \sum _{k=1}^{10}k=1+2+3+4+5+6+7+8+9+10}$

Now, we can add numbers.

  ${\displaystyle \sum _{k=1}^{10}k=55}$

So, the values of both sums are 55.

Conclusion:

Hence, both the sums are equal.

To determine

To check: whether the given sums ${\displaystyle \sum _{i=1}^{10}i}$ and ${\displaystyle \sum _{i=6}^{15}(i-5)}$ are equal or unequal.

Answer to Problem 1E

Both the sums ${\displaystyle \sum _{i=1}^{10}i}$ and ${\displaystyle \sum _{i=6}^{15}(i-5)}$ are equal.

Explanation of Solution

Given information:

The given sums are ${\displaystyle \sum _{i=1}^{10}i}$ and ${\displaystyle \sum _{i=6}^{15}(i-5)}$.

Concept used:

Summation expansion formula:

  ${\displaystyle \sum _{i=1}^n{a}_i=a_1+}{a}_2+a_3+a_4+\mathrm{.........}+a_n$

Calculation:

We will expand both sums.

  ${\displaystyle \sum _{i=1}^{10}i=1+2+3+4+5+6+7+8+9+10}$

Now, we can add numbers.

  ${\displaystyle \sum _{i=1}^{10}i=55}$

  ${\displaystyle \sum _{i=6}^{15}(i-5)}=\left[\begin{array}{l}(6-5)+(7-5)+(8-5)+(9-5)+(10-5)+\\ (11-5)+(12-5)+(13-5)+(14-5)+(15-5)\end{array}\right]$

Now, we can add numbers.

  ${\displaystyle \sum _{i=6}^{15}(i-5)}=55$

So, the values of both sums are 55.

Conclusion:

Hence, both the sums are equal.

To determine

To check: whether the given sums ${\displaystyle \sum _{i=1}^{10}i(i-1)}$ and ${\displaystyle \sum _{j=0}^9(j+1)j}$ are equal or unequal.

Answer to Problem 1E

Both the sums ${\displaystyle \sum _{i=1}^{10}i(i-1)}$ and ${\displaystyle \sum _{j=0}^9(j+1)j}$ are equal.

Explanation of Solution

Given information:

The given sums are ${\displaystyle \sum _{i=1}^{10}i(i-1)}$ and ${\displaystyle \sum _{j=0}^9(j+1)j}$ .

Concept used:

Summation expansion formula:

  ${\displaystyle \sum _{i=1}^n{a}_i=a_1+}{a}_2+a_3+a_4+\mathrm{.........}+a_n$

Calculation:

We will expand both sums.

  ${\displaystyle \sum _{i=1}^{10}i(i-1)}=\left[\begin{array}{l}1(1-1)+2(2-1)+3(3-1)+4(4-1)+5(5-1)+\\ 6(6-1)+7(7-1)+8(8-1)+9(9-1)+10(10-1)\end{array}\right]$

Now, we can add numbers.

  ${\displaystyle \sum _{i=1}^{10}i(i-1)}=330$

  ${\displaystyle \sum _{j=0}^9(j+1)j}=\left[\begin{array}{l}(0+1)0+(1+1)1+(2+1)2+(3+1)3+(4+1)4+\\ (5+1)5+(6+1)6+(7+1)7+(8+1)8+(9+1)9\end{array}\right]$

Now, we can add numbers.

  ${\displaystyle \sum _{j=0}^9(j+1)j}=330$

So, the values of both sums are 330.

Conclusion:

Hence, both the sums are equal.

To determine

To check: whether the given sums ${\displaystyle \sum _{i=1}^{10}i(i-1)}$ and ${\displaystyle \sum _{k=1}^{10}(k^2-k)}$ are equal or unequal.

Answer to Problem 1E

Both the sums ${\displaystyle \sum _{i=1}^{10}i(i-1)}$ and ${\displaystyle \sum _{k=1}^{10}(k^2-k)}$ are equal.

Explanation of Solution

Given information:

The given sums are ${\displaystyle \sum _{i=1}^{10}i(i-1)}$ and ${\displaystyle \sum _{k=1}^{10}(k^2-k)}$ .

Concept used:

Summation expansion formula:

  ${\displaystyle \sum _{i=1}^n{a}_i=a_1+}{a}_2+a_3+a_4+\mathrm{.........}+a_n$

Calculation:

We will expand both sums.

  ${\displaystyle \sum _{i=1}^{10}i(i-1)}=\left[\begin{array}{l}1(1-1)+2(2-1)+3(3-1)+4(4-1)+5(5-1)+\\ 6(6-1)+7(7-1)+8(8-1)+9(9-1)+10(10-1)\end{array}\right]$

Now, we can add numbers.

  ${\displaystyle \sum _{i=1}^{10}i(i-1)}=330$

  ${\displaystyle \sum _{k=1}^{10}(k^2-k)}=\left[\begin{array}{l}(1^2-1)+(2^2-2)+(3^2-3)+(4^2-4)+(5^2-5)+\\ (6^2-6)+(7^2-7)+(8^2-8)+(9^2-9)+({10}^2-10)\end{array}\right]$

Now, we can add numbers.

  ${\displaystyle \sum _{k=1}^{10}(k^2-k)}=330$

So, the values of both sums are 330.

Conclusion:

Hence, both the sums are equal.