Question

Statement 1: The sum of the series$1+(1+2+4)+(4+6+9)+(9+12+16)+\dots +(361+380+400)$ is $8000.$
Statement 2: $\sum _{k=1}^n(k^3-(k-1{)}^3)=n^3$ , for any natural number $n$.

• A. Statement 1 is false, Statement 2 is true.
• B. Statement 1 is true, Statement 2 is true; Statement 2 is a correct explanation for Statement 1.
• C. Statement 1 is true, Statement 2 is true; Statement 2 is not a correct explanation for Statement 1.
• D. Statement 1 is true , Statement 2 is false.

Answer 1

The correct option is B Statement 1 is true, Statement 2 is true; Statement 2 is a correct explanation for Statement 1.

$T_k=(k-1{)}^2+k(k-1)+k^2=\frac{\left(\right(k-1{)}^3-k^3)}{(k-1)-k}=k^3-(k-1{)}^3$
$S_k={\sum }_{k=1}^n{k}^3-(k^3-1-3k(k-1\left)\right)={\sum }_{k=1}^{n}3k^2-3k+1$

$=\frac{3n(n+1)(2n+1)}{6}-\frac{3k(k+1)}{2}+n=n^3$
$S_{20}={20}^3=8000$