Question

If the series of natural numbers is divided into groups (1), (2,3,4,5,6,7,8,9), (10,11,12,13,14,......,36), ..... and so on, then -

• A. total number of natural numbers in first n groups is $(\frac{n(n+1)}{2})^2$
• B. first term in $n^{th}$ group is $\frac{1}{4}(n-1)^2n^2 + 1$
• C. last term in $n^{th}$ group is $\frac{1}{4}(n-1)^2n^2 + n^3$
• D. the sum of the numbers in $n^{th}$ group is $\frac{n^3}{4}(n^4 + n^2 + 2)$

Answer 1

Answer: (A), (B), (C), (D)

All options (A), (B), (C), and (D) are correct.

Answer 2

1. The grouping is as follows:

   • Group 1 has 1 term = $1^3$.
   • Group 2 has 8 terms = $2^3$.
   • Group 3 has 27 terms = $3^3$.

2. Hence, the total number of natural numbers in the first $n$ groups is

   $$1^3+2^3+\cdots+n^3 = \left(\frac{n(n+1)}{2}\right)^2.$$

   This verifies (A).

3. For the first term of the $n^{th}$ group:

   • Since the first term of group 1 is 1 and that of group $n$ is one more than the last term of group $n-1$, by summing the count of numbers up to group $n-1$ we get $$\text{First term} = 1 + \sum_{k=1}^{n-1} k^3 = 1 + \left(\frac{(n-1)n}{2}\right)^2.$$
   • This is equivalent to $$\frac{(n-1)^2n^2}{4}+1,$$ which verifies (B).

4. The last term of the $n^{th}$ group is just the first term plus $n^3-1$. Therefore,

   $$\text{Last term} = \left[\frac{(n-1)^2n^2}{4}+1\right] + (n^3-1) = \frac{(n-1)^2n^2}{4}+n^3.$$

   This verifies (C).

5. The sum of the $n^{th}$ group is the sum of an arithmetic progression with:

   • First term: $\displaystyle a = \frac{(n-1)^2n^2}{4}+1$,
   • Last term: $\displaystyle l = \frac{(n-1)^2n^2}{4}+n^3$,
   • Number of terms: $n^3$.

   The sum is:

   $$S = \frac{n^3}{2}\left(a + l\right) = \frac{n^3}{2}\left[\frac{(n-1)^2n^2}{4}+1 + \frac{(n-1)^2n^2}{4}+n^3\right].$$

   Simplify the inside:

   $$= \frac{n^3}{2}\left[\frac{(n-1)^2n^2}{2}+ n^3+1\right].$$

   With further algebra (or verifying by testing small values $n=1,2,3$), one obtains

   $$\frac{n^3}{4}(n^4+n^2+2),$$

   which verifies (D).

Final Answer: All options (A), (B), (C), and (D) are correct.

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Explanation (Minimal):

• The total number of terms in first $n$ groups is the sum of cubes: $\sum_{k=1}^{n} k^3 = \left(\frac{n(n+1)}{2}\right)^2$.
• The first term in the $n^{th}$ group is one more than the sum of cubes up to $n-1$: $\frac{(n-1)^2n^2}{4}+1$.
• The last term in the $n^{th}$ group adds $n^3-1$ to the first term, giving $\frac{(n-1)^2n^2}{4}+n^3$.
• Using the AP sum formula, the group sum simplifies to $\frac{n^3}{4}(n^4+n^2+2)$.