Problem 155 - A Complete Resource Book in Mathematics for JEE Main

Question

Statement 1: The sum of the series \(1+(1+2+4)+\) \((4+6+9)+(9+12+16)+\ldots . .+(361+380+400)\) is 8000 . Statement 2: \(\sum_{k=1}^{n}\left(k^{3}=(k-1)^{3}\right)=n^{3}\) 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

Step-by-Step Solution

Answer

(A) Statement 1 is false, statement 2 is true.

Step 1: Identify the series in Statement 1

In Statement 1, observe the structure of the series given: \(1+(1+2+4)+(4+6+9)+(9+12+16)+ \ldots + (361+380+400)\). Notice the groups like \((1+2+4)\), \((4+6+9)\), \((9+12+16)\), each consisting of terms that are quadratic numbers. For example, \((1+2+4)\) corresponds to terms from quadratic numbers 1, 4, and 9.

Step 2: Analyze Statement 1's series further

Each group can be represented with terms in the form of \(a(a+1)/2\). The terms 1, 2, 4 are sums \(1^2\), \(2^2\), and \(3^2\) components. Similarly, other groups form successively: \((4=2^2,6=3^2+1,9=3^2)\), \((9=3^2,12=4^2,16=4^2)\), etc., reaching until \((400=20^2)\). Calculate the total sums of all terms in the sequence form.

Step 3: Calculate the total sum based on decomposed series

Decompose the series into a calculable structure by realizing that groups are complete squares addition: \((1+2+\ldots+20^2)\). Sum up using the formula for sum of square numbers \(1^2+2^2+3^2+\ldots+n^2 = \frac{n(n+1)(2n+1)}{6}\). For \(n=20\), calculate the sum.

Step 4: Evaluate Statement 1's sum

After evaluating the formula \(\sum_{k=1}^{20} k^2 = \frac{20 \times 21 \times 41}{6} = 2870\), we see Statement 1 claims a sum of 8000 is false, as we've calculated 2870.

Step 5: Evaluate Statement 2

The second statement, \(\sum_{k=1}^{n}(k^3-(k-1)^3)=n^3\), simplifies as each subtraction reduces the cubes effectively to \(n^3\). This statement is true as shown by expanding \((k^3-(k-1)^3 = 3k^2 - 3k + 1)\), which telescopically sums to \(n^3\).

Step 6: Determine correct option based on evaluations

Statement 1 is false because the series sum is not 8000, but Statement 2 is true, aligning with the identity. Therefore, the correct option is (A), as evaluated.

Key Concepts

Quadratic Numbers

Quadratic numbers are essential when studying sequences and series. A quadratic number is essentially a perfect square; it stems from squaring a whole number. For example, 1, 4, 9, and 16 are considered quadratic numbers because they result from 1², 2², 3², and 4² respectively.
When examining sequences that involve quadratic numbers, it's vital to recognize the pattern. In our exercise, the series includes sums of terms that build on these quadratic numbers, like (1+2+4), (4+6+9). Each set of terms can be expressed as sums of quadratic numbers (e.g., 1², 2²) progressing in a notable pattern.
Understanding how these numbers are formed is crucial for identifying sequences and finding sums of series.

Sum of Square Numbers

A significant concept in this problem involves summing square numbers. The squares of integers can be summed using the formula:

• \[\sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6}\]

This formula is powerful for quickly calculating the total sum of squared numbers up to any integer \(n\). For the exercise, determining the correct sum was crucial in verifying Statement 1. Here, the last term is square of 20, thus \(n = 20\).
Applying the formula:

• \[\frac{20 \cdot 21 \cdot 41}{6} = 2870\]

This result clearly shows that the sum of the stated series is indeed 2870, confirming the statement regarding its total as false as compared to the claimed 8000.

Cubic Numbers

Cubic numbers play an interesting role in sequences. A cubic number results from raising a whole number to the power of three. Examples include 1, 8, 27, which are 1³, 2³, and 3³.
The exercise's second statement focuses on solving differences of cubes. It asserts that:

• \[\sum_{k=1}^{n} (k^3 - (k-1)^3) = n^3\]

This simplifies through telescoping, where most terms cancel out, leaving \( n^3 \) as the final result. This property is crucial in resolving series of cubes effectively.
Understanding cubic numbers and their manipulation allows one to work with complex sequences more competently, as we observe here. In our example, exploring the structure of cubic equations reveals why Statement 2 is valid and correctly explains the algebraic manipulation provided in the solution.