Chapter 12.1, Problem 50E

To determine

Calculate the sum of the equation.

Answer to Problem 50E

The summation of the equation is $15550$ .

Explanation of Solution

Given: It is given in the question that the equation is ${\displaystyle \sum _{k=1}^{100}(3k+4)}$ .

Concept Used:

In this, use the concept of putting the value of variable in the equation and adding it and get the answer and in some case use the concept of nth term of sum of the series. The formula used is Sum of Arithmetic sequence : $\frac{n}{2}(firstterm+lastterm)$ .

Calculation: Here, the equation is ${\displaystyle \sum _{k=1}^{100}(3k+4)}$ .

  $\begin{array}{l}{\displaystyle \sum _{k=1}^{100}(3k+4)}=(3\cdot 1+4)+(3\cdot 2+4)+(3\cdot 3+4)+\mathrm{.............}+(3\cdot 99+4)+(3\cdot 100+4)\\ {\displaystyle \sum _{k=1}^{100}(3k+4)}=(3+4)+(6+4)+(9+4)+\mathrm{........}+(297+4)+(300+4)\\ {\displaystyle \sum _{k=1}^{100}(3k+4)}=7+10+13+\mathrm{.........}+301+304\end{array}$

Now, it is seen that the sequence that got $(7,10,13,\mathrm{.............},301,304)$ are arithmetic sequence. So use the formula to find the sum of arithmetic sequences.

Common difference : ( d ) = $a_n-a_{n-1}=10-7=3$

No of terms : $n=100$ , First term : $a_1=7$ ,Last term : $a_{100}=304$

Sum of Arithmetic sequence : $\frac{n}{2}(firstterm+lastterm)$

  $\begin{array}{l}{\displaystyle \sum _{k=1}^{100}(3k+4)}=\frac{100}{2}(7+304)\\ {\displaystyle \sum _{k=1}^{100}(3k+4)}=50\times 311=15550\end{array}$

Conclusion:

The summation is $15550$ .