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For the product n(n+1)(2n+1),n∈N which one of the following is not necessarily true ?
  • a)
    It is even
  • b)
    Divisible by 3
  • c)
    Divisible by the sum of the squares of first n natural numbers
  • d)
    Never divisible by 237
  • e)
    None of these
Correct answer is option 'D'. Can you explain this answer?
Verified Answer
For the productn(n+1)(2n+1),n∈N which one of the following is not...
Clearly the expression is basically the formula for sum of 1st n squares but without the "6" in the denominator. 
So options 1,2,4 are true.
And about 237  
The above expression will obviously be divisible by 237 
When n= 237 as it will be 237*238*475 In fact even for n = 236 and n = 118 it will be divisible by 237.



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Most Upvoted Answer
For the productn(n+1)(2n+1),n∈N which one of the following is not...
We can start by expanding the expression:

n(n 1)(2n 1) = 2n^3 + 3n^2 + n

Now, we can use the summation formulas:

∑n = (n(n+1))/2
∑n^2 = (n(n+1)(2n+1))/6

Using these formulas, we can write:

∑n(n 1)(2n 1) = ∑(2n^3 + 3n^2 + n)
= 2∑n^3 + 3∑n^2 + ∑n
= 2[(n(n+1)/2)^2] + 3[(n(n+1)(2n+1))/6] + (n(n+1))/2
= (n^2(n+1)^2)/2 + n(n+1)(2n+1)/2 + n(n+1)/2
= (n(n+1)/2)[n(n+1) + 2(2n+1) + 1]
= (n(n+1)/2)(2n^2 + 5n + 2)
= (n(n+1)(2n+1)(n+2))/3

Therefore, ∑n(n 1)(2n 1) = (n(n+1)(2n+1)(n+2))/3 for any positive integer n.
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For the productn(n+1)(2n+1),n∈N which one of the following is not necessarily true ?a)It is evenb)Divisible by 3c)Divisible by the sum of the squares of first n natural numbersd)Never divisible by 237e)None of theseCorrect answer is option 'D'. Can you explain this answer?
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