Answer:
The given expanded sum of the series is
![\sum\limits_(n=5)^(9)3n+2=115](https://img.qammunity.org/2021/formulas/mathematics/middle-school/a8khcjhd0xd6pmn5qolr4dqjojovder02c.png)
Explanation:
Given problem can be written as
![\sum\limits_(n=5)^(9)3n+2](https://img.qammunity.org/2021/formulas/mathematics/middle-school/dr9ve1uzl2v1t9xgzw4xnwo9c5gabafdt2.png)
To find their sums:
Now expanding the series
That is put n=5,6,7,8,9 in the given summation
![\sum\limits_(n=5)^(9)3n+2=[3(5)+2]+[3(6)+2]+[3(7)+2]+[3(8)+2]+[3(9)+2]](https://img.qammunity.org/2021/formulas/mathematics/middle-school/y8l1u4zz42qvsynatigwhklrcjujxyz2wg.png)
![=[15+2]+[18+2]+[21+2]+[24+2]+[27+2]](https://img.qammunity.org/2021/formulas/mathematics/middle-school/upsbmww1u2ci5xiou6w3sycvi3kldpnc8g.png)
(adding the terms)
![=115](https://img.qammunity.org/2021/formulas/mathematics/middle-school/vfvtfpbmtpd5tuae2zrxbjan88wj4jvxip.png)
Therefore
![\sum\limits_(n=5)^(9)3n+2=115](https://img.qammunity.org/2021/formulas/mathematics/middle-school/a8khcjhd0xd6pmn5qolr4dqjojovder02c.png)
Therefore the given sum of the series is
![\sum\limits_(n=5)^(9)3n+2=115](https://img.qammunity.org/2021/formulas/mathematics/middle-school/a8khcjhd0xd6pmn5qolr4dqjojovder02c.png)
The given expanded sum of the series is
![\sum\limits_(n=5)^(9)3n+2=115](https://img.qammunity.org/2021/formulas/mathematics/middle-school/a8khcjhd0xd6pmn5qolr4dqjojovder02c.png)