Question

find the value

Answer 1

\[\sum_{r=16}^{30}(r+2)(r-3)\]

Answer 2

hint:
we can rewrite your sum as below:
\[\large \sum\limits_{r = 16}^{30} {{r^2} - r - 6} = \left( {\sum\limits_{r = 16}^{30} {{r^2} - r} } \right) - 6 \times \left( {30 - 16 + 1} \right) = ...?\]

Answer 3

well
i dont have any idea about this
i just got this question
i need to solve this :)
if u could just help me to understand the basics:)

Answer 4

ok!

Answer 5

step #1
I rewrote your expression as below:
\[\large \left( {r + 2} \right)\left( {r - 3} \right) = {r^2} - r - 6\]

Answer 6

i got it
next:)

Answer 7

i meant i got ur step 1

Answer 8

so I have broken your sum in other three sums, as below:

\[\large \sum\limits_{r = 12}^{30} {\left( {{r^2} - r - 6} \right)} = \left( {\sum\limits_{r = 12}^{30} {{r^2}} } \right) - \left( {\sum\limits_{r = 12}^{30} r } \right) - \left( {\sum\limits_{r = 12}^{30} 6 } \right)\]

Answer 9

well r=16
got that

Answer 10

oops.. the minimum value of r is not 12, it is 16

Answer 11

now, the third sum is easy to solve. Its value is:
\[\large \sum\limits_{r = 16}^{30} 6 = 6 \times \left( {30 - 16 + 1} \right) = ...?\]

Answer 12

the value is \[6\ X\ (30-16+1)=6\ X\ 15 =90\]

Answer 13

ok!

Answer 14

the second sum is the sum of an arithmetic sequence, whose constant is 1. Its sum is given by the subsequent formula:
\[\large \frac{{{a_{16}} + {a_{30}}}}{2} \times \left( {30 - 16 + 1} \right) = ...?\]

Answer 15

where:
\[\large \begin{gathered}
{a_{16}} = 16 \hfill \\
{a_{30}} = 30 \hfill \\
\end{gathered} \]

Answer 16

so, the second sum is:
\[\large \frac{{16 + 30}}{2} \times \left( {30 - 16 + 1} \right) = ...?\]

Answer 17

the answer is 345

Answer 18

ok!

Answer 19

Now, I can rewrite the third sum as below:
\[\large \sum\limits_{r = 16}^{30} {{r^2}} = \sum\limits_{r = 1}^{30} {{r^2}} - \sum\limits_{r = 1}^{15} {{r^2}} \]

Answer 20

oops.. the first sum

Answer 21

i did not get that :(

Answer 22

I have broken the first sum as above, since I want to apply this formula:
\[\large \sum\limits_{r = 1}^n {{r^2}} = \frac{{n\left( {n + 1} \right)\left( {2n + 1} \right)}}{6}\]

Answer 23

that for mula is general, and it can be proved using the induction principle of mathematics

Answer 24

okay :)

Answer 25

so, our first sum is:
\[\large \begin{gathered}
\sum\limits_{r = 16}^{30} {{r^2}} = \sum\limits_{r = 1}^{30} {{r^2}} - \sum\limits_{r = 1}^{15} {{r^2}} = \hfill \\
\hfill \\
= \frac{{30\left( {30 + 1} \right)\left( {2 \times 30 + 1} \right)}}{6} - \frac{{15\left( {15 + 1} \right)\left( {2 \times 15 + 1} \right)}}{6} = ...? \hfill \\
\end{gathered} \]

Answer 26

The answer is 8215,right?

Answer 27

ok!

Answer 28

so, your answer, is:
8215-345-90=...?

Answer 29

now adding up all together we get
\[8215-345-90=7780\]

Answer 30

that's right!

Answer 31

Thank you so much @Michele_Laino
you are one of the best user !!!!!!!!

Answer 32

Thank you!!!!! :) @rvc