Consider the following statements : 1. 2 + 4 + 6 + ........ + 2n = n 2 + n 2. The expression n 2 + n + 41 always gives a prime number for every natural number n Which of the above statements is/are correct ?
1 only
Understanding the Mathematical Statements
The question asks us to evaluate the correctness of two given mathematical statements. We need to analyze each statement independently to determine if it is true for all conditions specified within the statement.
Analyzing Statement 1: Sum of First n Even Numbers
Statement 1 says: \(2 + 4 + 6 + ........ + 2n = n^2 + n\).
This statement represents the sum of the first $n$ positive even numbers. This sequence \((2, 4, 6, \dots, 2n)\) is an arithmetic progression (AP) because the difference between consecutive terms is constant.
- First term ($a$) = 2
- Common difference ($d$) = 4 - 2 = 2
- Number of terms = $n$
- Last term ($l$) = $2n$
The sum of an arithmetic progression (\(S_n\)) can be calculated using the formula: \(S_n = \frac{n}{2} [2a + (n-1)d]\) Let's substitute the values of $a$ and $d$ into the formula: \(S_n = \frac{n}{2} [2(2) + (n-1)2]\) \(S_n = \frac{n}{2} [4 + 2n - 2]\) \(S_n = \frac{n}{2} [2n + 2]\) Now, we can factor out 2 from the term inside the square brackets: \(S_n = \frac{n}{2} [2(n + 1)]\) \(S_n = n(n + 1)\) \(S_n = n^2 + n\)
The calculated sum of the first $n$ even numbers is \(n^2 + n\). This matches the expression given in Statement 1. Therefore, Statement 1 is correct.
Analyzing Statement 2: Prime Generating Expression
Statement 2 says: The expression \(n^2 + n + 41\) always gives a prime number for every natural number $n$.
A prime number is a natural number greater than 1 that has no positive divisors other than 1 and itself. Natural numbers are 1, 2, 3, 4, and so on.
The statement claims that for every natural number $n$, the expression \(n^2 + n + 41\) will result in a prime number. To check this, we can test the expression for some values of $n$.
- For $n=1$: \(1^2 + 1 + 41 = 1 + 1 + 41 = 43\). 43 is a prime number.
- For $n=2$: \(2^2 + 2 + 41 = 4 + 2 + 41 = 47\). 47 is a prime number.
- For $n=3$: \(3^2 + 3 + 41 = 9 + 3 + 41 = 53\). 53 is a prime number.
The expression gives prime numbers for these initial values. However, the statement says "always" for "every natural number n". If we can find just one natural number $n$ for which the expression does not produce a prime number, the statement is false.
Let's consider the value $n=40$.
- For $n=40$: \(40^2 + 40 + 41\).
We can calculate this value: \(40^2 + 40 + 41 = 1600 + 40 + 41 = 1681\) Now we need to check if 1681 is a prime number. Consider the original expression for $n=40$: \(40^2 + 40 + 41\). We can factor this expression: \(40^2 + 40 + 41 = 40(40 + 1) + 41 = 40(41) + 41\) Now we can factor out 41: \(40(41) + 41 = 41(40 + 1) = 41 \times 41\) So, for $n=40$, the expression evaluates to \(41 \times 41 = 1681\). Since 1681 has a divisor 41 (which is not 1 or 1681), it is not a prime number; it is a composite number.
Since the expression \(n^2 + n + 41\) does not give a prime number for $n=40$ (or for $n=41$ where it gives \(41^2 + 41 + 41 = 41(41+1+1) = 41 \times 43\)), Statement 2 is incorrect.
Comparing the Statements
Based on our analysis:
- Statement 1: \(2 + 4 + 6 + ........ + 2n = n^2 + n\) is Correct.
- Statement 2: The expression \(n^2 + n + 41\) always gives a prime number for every natural number $n$ is Incorrect.
Therefore, only Statement 1 is correct.
Revision Table: Summary of Statement Analysis
| Statement Number | Statement Description | Correctness |
|---|---|---|
| 1 | Sum of first n even numbers equals \(n^2 + n\) | Correct |
| 2 | Expression \(n^2 + n + 41\) always gives a prime for natural number n | Incorrect |
Additional Information: Key Mathematical Concepts
Let's briefly touch upon some related concepts.
Arithmetic Series
An arithmetic series is the sum of the terms of an arithmetic sequence. A sequence is arithmetic if the difference between consecutive terms is constant. The formula for the sum of the first $n$ terms is \(S_n = \frac{n}{2}(a_1 + a_n)\) or \(S_n = \frac{n}{2}(2a_1 + (n-1)d)\), where \(a_1\) is the first term, \(a_n\) is the nth term, and $d$ is the common difference. In Statement 1, the sequence of even numbers starting from 2 (\(2, 4, 6, \dots\)) is an arithmetic sequence.
Prime Numbers and Composite Numbers
A prime number is a natural number greater than 1 that has exactly two distinct positive divisors: 1 and itself (e.g., 2, 3, 5, 7, 11). A composite number is a natural number greater than 1 that is not prime; it has more than two positive divisors (e.g., 4, 6, 8, 9, 10). The number 1 is neither prime nor composite.
Euler's Prime-Generating Polynomial
The expression \(n^2 + n + 41\) is a famous example, often referred to in the context of prime numbers. It generates primes for many initial values of $n$ (specifically for \(n = 0, 1, 2, \dots, 39\)). However, as demonstrated, it fails for $n=40$, $n=41$, and many subsequent values. This shows that an expression generating many primes initially does not necessarily generate primes infinitely or always for every natural number.
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