Question:
Which of the following statements are true? A. \(2^{4n}-1\) is divisible by \(15\).
B. \(5,12\) and \(13\) is Pythagorean triplet.
C. \(n^7-n\) is divisible by \(42\).
D. \(1^2+2^2+3^2+\cdots+n^2=\left[\dfrac{n(n+1){2}\right]^2\).}
Which of the following statements are true? A. \(2^{4n}-1\) is divisible by \(15\).
B. \(5,12\) and \(13\) is Pythagorean triplet.
C. \(n^7-n\) is divisible by \(42\).
D. \(1^2+2^2+3^2+\cdots+n^2=\left[\dfrac{n(n+1){2}\right]^2\).}
B. \(5,12\) and \(13\) is Pythagorean triplet.
C. \(n^7-n\) is divisible by \(42\).
D. \(1^2+2^2+3^2+\cdots+n^2=\left[\dfrac{n(n+1){2}\right]^2\).}
Show Hint
Remember: sum of squares is \(\frac{n(n+1)(2n+1)}{6}\), while sum of cubes is \(\left[\frac{n(n+1)}{2}\right]^2\).
Updated On: Jun 6, 2026
- A and B only
- A, B and D only
- A, B and C only
- A, C and D only
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The Correct Option is C
Solution and Explanation
Concept:
We check each mathematical statement separately.
Step 1: Check statement A. \[ 2^{4n}-1=(2^4)^n-1=16^n-1 \] Since, \[ 16\equiv 1 \pmod{15} \] Therefore, \[ 16^n\equiv 1^n \pmod{15} \] \[ 16^n-1\equiv 0 \pmod{15} \] So, A is true.
Step 2: Check statement B. \[ 5^2+12^2=25+144=169 \] \[ 13^2=169 \] So, \[ 5^2+12^2=13^2 \] Hence, \(5,12,13\) is a Pythagorean triplet. B is true.
Step 3: Check statement C. \[ n^7-n \] This expression is divisible by: \[ 2,\quad 3,\quad 7 \] Therefore, it is divisible by: \[ 2\cdot 3\cdot 7=42 \] So, C is true.
Step 4: Check statement D.
The correct formula for sum of squares is: \[ 1^2+2^2+3^2+\cdots+n^2=\frac{n(n+1)(2n+1)}{6} \] But the given formula: \[ \left[\frac{n(n+1)}{2}\right]^2 \] is the formula for: \[ 1^3+2^3+3^3+\cdots+n^3 \] So, D is false. Thus, true statements are: \[ A,\ B,\ C \] \[ \therefore \text{Correct Answer is A, B and C only} \]
We check each mathematical statement separately.
Step 1: Check statement A. \[ 2^{4n}-1=(2^4)^n-1=16^n-1 \] Since, \[ 16\equiv 1 \pmod{15} \] Therefore, \[ 16^n\equiv 1^n \pmod{15} \] \[ 16^n-1\equiv 0 \pmod{15} \] So, A is true.
Step 2: Check statement B. \[ 5^2+12^2=25+144=169 \] \[ 13^2=169 \] So, \[ 5^2+12^2=13^2 \] Hence, \(5,12,13\) is a Pythagorean triplet. B is true.
Step 3: Check statement C. \[ n^7-n \] This expression is divisible by: \[ 2,\quad 3,\quad 7 \] Therefore, it is divisible by: \[ 2\cdot 3\cdot 7=42 \] So, C is true.
Step 4: Check statement D.
The correct formula for sum of squares is: \[ 1^2+2^2+3^2+\cdots+n^2=\frac{n(n+1)(2n+1)}{6} \] But the given formula: \[ \left[\frac{n(n+1)}{2}\right]^2 \] is the formula for: \[ 1^3+2^3+3^3+\cdots+n^3 \] So, D is false. Thus, true statements are: \[ A,\ B,\ C \] \[ \therefore \text{Correct Answer is A, B and C only} \]
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