Sequences & Series
Sum of Cubes
Grade 11

Question:

<p><strong>Statement-1:</strong> The sum of the series \(1 + (1 + 2 + 4) + (4 + 6 + 9) + (9 + 12 + 16) + \ldots + (361 + 380 + 400)\) is 8000.</p><p><strong>Statement-2:</strong> For any natural number \(n\), \(\sum_{k=1}^{n} k^3 = \left[\frac{n(n+1)}{2}\right]^2\)</p>
<p>(A) Statement-1 is true, Statement-2 is true, Statement-2 is a correct explanation for Statement-1</p>
<p>(B) Statement-1 is true, Statement-2 is true, Statement-2 is NOT the correct explanation for Statement-1</p>
<p>(C) Statement-1 is true, Statement-2 is false</p>
<p>(D) Statement-1 is false, Statement-2 is true</p>

Step-by-Step Solution

Key Concept: Recognize that the grouped series represents sum of cubes and use the identity \(\sum k^3 = [n(n+1)/2]^2\) to verify.
**Step 1: Verify Statement-1.** The given series is $S = 1 + (1 + 2 + 4) + (4 + 6 + 9) + (9 + 12 + 16) + \ldots + (361 + 380 + 400)$. The first term is $1$. The subsequent groups follow a pattern. Let's examine the $m$-th group (for $m \ge 2$). The terms in the $m$-th group are $( (m-1)^2, (m-1)m, m^2 )$. For example: For $m=2$: $(1^2, 1 \cdot 2, 2^2) = (1, 2, 4)$. For $m=3$: $(2^2, 2 \cdot 3, 3^2) = (4, 6, 9)$. The last group is $(361, 380, 400)$, which corresponds to $(19^2, 19 \cdot 20, 20^2)$. Thus, the index $m$ for the groups runs from $2$ to $20$. The sum of the terms within each group $(m-1)^2 + (m-1)m + m^2$ can be simplified: $(m-1)^2 + (m-1)m + m^2 = (m^2 - 2m + 1) + (m^2 - m) + m^2 = 3m^2 - 3m + 1$. This expression is also equivalent to $m^3 - (m-1)^3$. Therefore, the series can be written as: $S = 1 + \sum_{m=2}^{20} (m^3 - (m-1)^3)$. This is a telescoping sum: $S = 1 + [(2^3 - 1^3) + (3^3 - 2^3) + \ldots + (20^3 - 19^3)]$. $S = 1 + (20^3 - 1^3)$. $S = 1 + (8000 - 1)$. $S = 8000$. Thus, Statement-1 is true. **Step 2: Verify Statement-2.** Statement-2 asserts that for any natural number $n$, $\sum_{k=1}^{n} k^3 = \left[\frac{n(n+1)}{2}\right]^2$. This is a fundamental identity for the sum of the first $n$ cubes. Thus, Statement-2 is true. **Step 3: Determine if Statement-2 is a correct explanation for Statement-1.** The sum of the series in Statement-1 is $8000$, which is $20^3$. The value $20^3$ can be expressed as the difference of two consecutive sums of cubes: $20^3 = \sum_{k=1}^{20} k^3 - \sum_{k=1}^{19} k^3$. Using the identity from Statement-2, we can calculate each sum: $\sum_{k=1}^{20} k^3 = \left[\frac{20(20+1)}{2}\right]^2 = \left[\frac{20 \cdot 21}{2}\right]^2 = (10 \cdot 21)^2 = (210)^2 = 44100$. $\sum_{k=1}^{19} k^3 = \left[\frac{19(19+1)}{2}\right]^2 = \left[\frac{19 \cdot 20}{2}\right]^2 = (19 \cdot 10)^2 = (190)^2 = 36100$. Substituting these values into the expression for $20^3$: $20^3 = 44100 - 36100 = 8000$. This demonstrates that the value of the series in Statement-1 can be derived using the identity provided in Statement-2. Therefore, Statement-2 is a correct explanation for Statement-1.
Correct Answer: A

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