Question

Please help me in this question​

Answer 1

Step-by-step explanation:

$\color{olive}\[ \begin{array}{l} \tt \: consider \: \sum_{r=1}^{10}\left(2r - 1\right) \\ \\ \tt=\sum_{r=1}^{10}\left(4 r^{2}+1-4 r\right) \\ \\ \tt=\sum_{r=1}^{10} 4 r^{2}-\sum_{r=1}^{10} 4 r+\sum_{r=1}^{10} 1 \\ \\ \tt =4 \sum_{r=1}^{10} r^{2}-4 \sum_{r=1}^{10} r+\sum_{r=1}^{10} 1 \ldots \ldots .(1) \\ \\ \tt \text{We know that} \sum_{r=1}^{n} r^{2}= \[ \dfrac{n(n+1)(2 n+1)}{6} \] \\ \\ \[ \begin{array}{l} \tt\sum_{r=1}^{n} r=\dfrac{n(n+1)}{2} \\ \\ \text { and } \tt \sum_{r=1}^{n} 1=n \end{array} \] \\ \\ \text{ Using these results in equation (1) becomes: } \\ \\ \[ \begin{array}{l} \tt =4 \times \dfrac{10(11)(21)}{6}-4 \times \dfrac{10(11)}{2}+10 \\ \\ \tt=1540-220+10=1330 \end{array} \] \end{array} \]$

Answer 2

Answer:

Consider,

$\small \longrightarrow \sum \limits_{r = 1}^{10}(2r - 1)^{2}$

$\small \longrightarrow \sum \limits_{r = 1}^{10}(4 {r}^{2} + 1 - 4r)$

$\small \longrightarrow \sum \limits_{r = 1}^{10}(4 {r}^{2}) +\sum \limits_{r = 1}^{10}( 1) -\sum \limits_{r = 1}^{10}( 4r)$

${ \small \longrightarrow 4 \sum \limits_{r = 1}^{10}({r}^{2}) +10 -4\sum \limits_{r = 1}^{10}(r)}$

Now we know that:

• $\small{\sum\limits_{r=1}^n r = \dfrac{n(n+1)}{2}}$
• $\small{\sum\limits_{r=1}^n r^2= \dfrac{n(n+1)(2n+1)}{6}}$

By using these results, we get:

${ \small \longrightarrow 4 \left( \dfrac{10(10 + 1)(20 + 1)}{6} \right) +10 -4\left( \dfrac{10(10 + 1)}{2} \right)}$

${ \small \longrightarrow 4 \left( \dfrac{10(11)(21)}{6} \right) +10 -4\left( \dfrac{10(11)}{2} \right)}$

${ \small \longrightarrow 4( 5)(11)(7) +10 -4( 5)(11) }$

${ \small \longrightarrow 1540+10 -220}$

${ \small \longrightarrow 1330}$

So the required answer is:

$\underline{ \boxed{ \sum \limits_{r = 1}^{10}(2r - 1)^{2} = 1330}}$