Question

The value of $\sum\limits_{r=16}^{30}{\left( r+2 \right)\left( r-3 \right)}$ is equal to?
A.7782
B.7787
C.7790
D.7780

Answer 1

In order to evaluate the value of the given series, firstly we have to classify the given problem into the type of series it belongs to. Then we would be considering the given problem and correlate it with the general form of product of the sum of the series.

Complete step-by-step solution:
Now let us learn about the evaluating of the geometric series. Generally, to check whether the sequence is geometric, we can simply check it by the ratio of the successive entries in the series have the same ratio or not. The common ratio can also be negative which will result in an alternating sequence.
Now let us evaluate the value of the given terms i.e. $\sum\limits_{r=16}^{30}{\left( r+2 \right)\left( r-3 \right)}$.
Firstly, let us equate it to $S$, we get
$S=\sum\limits_{r=16}^{30}{\left( r+2 \right)\left( r-3 \right)}$.
In order to find the value, we will be expressing the given problem in such a way that would be easily expressed in the general forms.
Upon following this, we will be expressing in the following way.

& S=\sum\limits_{r=16}^{30}{\left( r+2 \right)\left( r-3 \right)} \\\ & \Rightarrow \sum\limits_{r=1}^{30}{\left( r+2 \right)\left( r-3 \right)}-\sum\limits_{r=1}^{15}{\left( r+2 \right)\left( r-3 \right)} \\\ \end{aligned}$$ Upon solving this, we get $$\Rightarrow \sum\limits_{r=1}^{30}{\left( {{r}^{2}}-r-6 \right)-}\sum\limits_{r=1}^{15}{\left( {{r}^{2}}-r-6 \right)}$$ Now we will be using the general formula i.e. $$\sum\limits_{i=1}^{n}{{{k}^{2}}={{1}^{2}}+{{2}^{2}}+{{3}^{2}}+....{{n}^{2}}=\dfrac{n\left( n+1 \right)\left( n+2 \right)}{6}}$$ Now we get, $$\Rightarrow $$$${{\left( \dfrac{r\left( r+1 \right)\left( 2r+1 \right)}{6}-\dfrac{r\left( r+1 \right)}{2}-6r \right)}_{r=30}}-{{\left( \dfrac{r\left( r+1 \right)\left( 2r+1 \right)}{6}-\dfrac{r\left( r+1 \right)}{2}-6r \right)}_{r=15}}$$ Now, upon evaluating by substituting the values, we obtain $$\Rightarrow \left( \dfrac{30\left( 30+1 \right)\left( 2\left( 30 \right)+1 \right)}{6}-\dfrac{30\left( 30+1 \right)}{2}-6\left( 30 \right) \right)-\left( \dfrac{15\left( 15+1 \right)\left( 2\left( 15 \right)+1 \right)}{6}-\dfrac{15\left( 15+1 \right)}{2}-6\left( 15 \right) \right)$$ Upon solving this, we get $$\Rightarrow \left( \dfrac{30\left( 31 \right)\left( 60+1 \right)}{6}-\dfrac{30\left( 31 \right)}{2}-180 \right)-\left( \dfrac{15\left( 16 \right)\left( 31 \right)}{6}-\dfrac{15\left( 16 \right)}{2}-90 \right)$$ On simplifying this, we get the value as $$\Rightarrow \left( \dfrac{30\left( 31 \right)\left( 60+1 \right)}{6}-\dfrac{30\left( 31 \right)}{2}-180 \right)-\left( \dfrac{15\left( 16 \right)\left( 31 \right)}{6}-\dfrac{15\left( 16 \right)}{2}-90 \right)=7780$$ **$$\therefore $$ The value of $$\sum\limits_{r=16}^{30}{\left( r+2 \right)\left( r-3 \right)}$$ is equal to $$7780$$.** **Note:** We can apply the geometric sequences in our everyday life, we can use it for calculating interest earned. In geometric sequence, in order to get from one term to the other, we will be multiplying unlike the arithmetic sequence in which we add to get the other terms.