Question

Statement - 1: The sum of the series 1 + (1 + 2 + 4) + (4 + 6 + 9) + (9 + 12 + 16)+...+(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.

$n^{th}$ term of the given series
$T_n=(n-1{)}^2+(n-1)n+n^2$
$=\frac{\left(\right(n-1{)}^3-n^3)}{(n-1)-n}=n^3-(n-1{)}^3\Rightarrow S_n={\sum }_{k=1}^n[k^3-(k-1{)}^3]\Rightarrow 8000=n^3$
$\Rightarrow n=20$ which is a natural number
$T_1=1^3-0^3{T}_2=2^3-1^3⋮T_{20}={20}^3-{19}^3$
Now, $T_1+T_2+...+T_{20}=S_{20}\Rightarrow S_{20}={20}^3-0^3=8000$
Hence, both the given statements are true and statement 2 supports statement 1.