Question

Statement I. The sum of the series $1+(1+2+4)+(4+6+9)+(9+12+16)+\dots +(361+380+400)$ is $8000$.

Statement II$\sum _{k=1}^n\left[k^3\right.\left.-(k-1{)}^3\right]=n^3$ , for any natural number $n.$

• A. Statement I is false, Statement II is true
• B. Statement I is true, Statement II is true; Statement II is a correct explanation of Statement I
• C. Statement I is true, Statement II is true; Statement II is not a correct explanation of Statement I
• D. Statement I is true, Statement II is false

Answer 1

The correct option is B

Statement I is true, Statement II is true; Statement II is a correct explanation of Statement I

Explanation for the correct option:

Finding the sum of the series in statement I and solving the given expression in statement II:

On expanding the given polynomial by placing values of $k=0,1,2,.........,n$

$$\begin{gathered} \sum _{k=1}^n\left[k^3-(k-1{)}^3\right]=n^3 \\ \Rightarrow 1^3-0^3+2^3-1^3+3^3-2^3+\dots n^3-(n-1{)}^3=n^3 \end{gathered}$$

Now solving for $n=20,$as the total number of digits in series are $20$.

Substituting $n=20,$ in the expression $\sum _{k=1}^n\left[k^3\right.\left.-(k-1{)}^3\right]=n^3$

$$\begin{gathered} \Rightarrow 1^3-0^3+2^3-1^3+3^3-2^3+\dots {20}^3-{19}^3={20}^3 \\ \Rightarrow 1^3+(2-1)(4+2+1)+(3-2)\left(3^2+2^2+6\right)+\dots .(20-19)\left(400+{19}^2+20\times 19\right)={20}^3 \\ \Rightarrow 1+(1+2+4)+(4+6+9)+\dots .\left(1\right)(400+361+380)=8000 \end{gathered}$$ [Using, $a^3-b^3=(a-b)\left(a^2+ab+b^2\right)$]

Therefore, Statement I and Statement II are true; and Statement II is a correct explanation of Statement I.

Therefore, the correct answer is option (B).