Question

Please find 10th & 11th​

Answer 1

Step-by-step explanation:

we have,

$\sum _{r = 1}^n(5 {r}^{2} + 4r - 3)$

$= \sum _{r = 1}^n(5 {r}^{2} ) + \sum _{r = 1}^n(4r) - \sum _{r = 1}^n(3)$

$= 5(\sum _{r = 1}^n {r}^{2}) + 4(\sum _{r = 1}^nr) - 3n$

$= \frac{5n(n + 1)(2n + 1)}{6} + \frac{4n(n + 1)}{2} - 3n$

$= \frac{5n(n + 1)(2n + 1) + 12n(n + 1) - 18n}{6}$

$= \frac{n(10n^{2} + 15n + 5 + 12n + 12 - 18n)}{6}$

$= \frac{n(10 {n}^{2} + 9n + 17) }{6}$

Answer 2

Answer:

10. n(10n²+27n-1)/6

i hope this is helpful for you