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n(n-1)(2n-1) is divisible by
- a)15
- b)6
- c)4
- d)64
Correct answer is option 'C'. Can you explain this answer?
Verified Answer
n(n-1)(2n-1) is divisible bya)15b)6c)4d)64Correct answer is option 'C'...
For n=1
n(n+1)(2n+1) = 6, divisible by 6.
Let the result be true for n=k
Then, k(k+1)(2k+1) is divisible by 6.
So k(k+1)(2k+1) =6m (1)
Now to prove that the result is true for n=k+1
That is to prove, (K+1)(k+2)(2k+3) is divisible by 6.
(K+1)(k+2)(2k+3)=(k+1)k(2k+3)+(k+1)2(2k+3)=(k+1)k(2k+1)+(k+1)k2+(k+1)2(2k+3)
=6m+2(k+1)(k+2k+3) using (1)
=6m+2(k+1)(3k+3)
=6m +6(k+1)(k+1)=6[m+(k+1)^2]
So divisible by6.
Most Upvoted Answer
n(n-1)(2n-1) is divisible bya)15b)6c)4d)64Correct answer is option 'C'...
Given expression: n(n-1)(2n-1)
We need to check whether this expression is divisible by 4 or not.
Solution:
We know that for a number to be divisible by 4, its last two digits should be divisible by 4.
Let's consider the given expression for different values of n:
For n = 1, the expression evaluates to 1(0)(1) = 0, which is divisible by 4.
For n = 2, the expression evaluates to 2(1)(3) = 6, which is not divisible by 4.
For n = 3, the expression evaluates to 3(2)(5) = 30, which is divisible by 4.
For n = 4, the expression evaluates to 4(3)(7) = 84, which is divisible by 4.
We can observe that the expression is divisible by 4 for every alternate value of n.
Hence, the answer is option 'C' (4).
We need to check whether this expression is divisible by 4 or not.
Solution:
We know that for a number to be divisible by 4, its last two digits should be divisible by 4.
Let's consider the given expression for different values of n:
For n = 1, the expression evaluates to 1(0)(1) = 0, which is divisible by 4.
For n = 2, the expression evaluates to 2(1)(3) = 6, which is not divisible by 4.
For n = 3, the expression evaluates to 3(2)(5) = 30, which is divisible by 4.
For n = 4, the expression evaluates to 4(3)(7) = 84, which is divisible by 4.
We can observe that the expression is divisible by 4 for every alternate value of n.
Hence, the answer is option 'C' (4).
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