Question

Class 11th
plz help me​

Answer 1

Answer:

$\red {\sum (r+1)(2r-1)} \green {=r\left(\frac{4r^{2}+9r-1}{6}\right)}$

Step-by-step explanation:

/* Upper boundary is not mentioned here .

$\sum (r+1)(2r-1) \\= \sum (2r^{2}-r+2r-1)$

$= \sum (2r^{2} + r - 1)\\= 2\sum r^{2} + \sum r - \sum 1$

$= 2\left(\frac{r(r+1)(2r+1)}{6}\right)+ \frac{r(r+1)}{2} - r$

$= r\left(\frac{(r+1)(2r+1)}{3} + \frac{r+1}{2} - 1 \right)\\= r\left(\frac{(2r^{2}+3r+1)}{3} + \frac{r+1}{2} - 1 \right)$

$=r\left( \frac{2(2r^{2}+3r+1)+3(r+1)-6)}{6}\right)\\=r\left( \frac{4r^{2}+6r+2+3r+3-6}{6}\right)\\= r\left(\frac{4r^{2}+9r-1}{6}\right)$

Therefore.,

$\red {\sum (r+1)(2r-1)} \green {=r\left(\frac{4r^{2}+9r-1}{6}\right)}$

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