Question

Answer it fast .....plsss....​

Answer 1

The following sum is given as assertion here.

$\longrightarrow\sf{(1)+(1+2+4)+(4+6+9)+\,\dots\,+(81+90+100)=1000}$

Each term in the series is formed as the following.

• $\sf{a_1=1=0^2+0\times1+1^2}$

• $\sf{a_2=1+2+4=1^2+1\times2+2^2}$

• $\sf{a_3=4+6+9=2^2+2\times3+3^2}$

Final term is,

• $\sf{a_{10}=81+90+100=9^2+9\times10+10^2}$

Our series contains 10 terms.

So $\sf{r^{th}}$ term of the series is,

$\longrightarrow\sf{a_r=(r-1)^2+r(r-1)+r^2}$

$\longrightarrow\sf{a_r=r^2-2r+1+r^2-r+r^2}$

$\longrightarrow\sf{a_r=3r^2-3r+1}$

$\longrightarrow\sf{a_r=r^3-r^3+3r^2-3r+1}$

$\longrightarrow\sf{a_r=r^3-(r^3-3r^2+3r-1)}$

$\longrightarrow\sf{a_r=r^3-(r-1)^3}$

Hence our sum becomes,

$\displaystyle\longrightarrow\sf{\sum_{r=1}^{10}\left(r^3-(r-1)^3\right)=1000}$

$\displaystyle\longrightarrow\sf{\sum_{r=1}^{10}\left(r^3-(r-1)^3\right)=10^3}$

This can imply, for $\sf{n\in\mathbb{N},}$

$\displaystyle\longrightarrow\sf{\sum_{r=1}^{n}\left(r^3-(r-1)^3\right)=n^3}$

Let us check if it's true.

$\displaystyle\longrightarrow\sf{\sum_{r=1}^{n}\left(r^3-(r-1)^3\right)=\sum_{r=1}^{n}\left(3r^2-3r+1\right)}$

$\displaystyle\longrightarrow\sf{\sum_{r=1}^{n}\left(r^3-(r-1)^3\right)=3\sum_{r=1}^{n}r^2-3\sum_{r=1}^{n}r+\sum_{r=1}^{n}1}$

$\displaystyle\longrightarrow\sf{\sum_{r=1}^{n}\left(r^3-(r-1)^3\right)=3\cdot\dfrac{n(n+1)(2n+1)}{6}-3\cdot\dfrac{n(n+1)}{2}+n}$

$\displaystyle\longrightarrow\sf{\sum_{r=1}^{n}\left(r^3-(r-1)^3\right)=\dfrac{n(n+1)(2n+1)-3n(n+1)+2n}{2}}$

$\displaystyle\longrightarrow\sf{\sum_{r=1}^{n}\left(r^3-(r-1)^3\right)=\dfrac{n[(n+1)(2n+1-3)+2]}{2}}$

$\displaystyle\longrightarrow\sf{\sum_{r=1}^{n}\left(r^3-(r-1)^3\right)=\dfrac{2n[(n+1)(n-1)+1]}{2}}$

$\displaystyle\longrightarrow\sf{\sum_{r=1}^{n}\left(r^3-(r-1)^3\right)=n(n^2-1+1)}$

$\displaystyle\longrightarrow\sf{\sum_{r=1}^{n}\left(r^3-(r-1)^3\right)=n^3}$

So it's true.

Therefore, both (A) and (R) are true, and (R) is correct explanation of (A).