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04 Mar 2020, 07:29
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For which of the following values of n , the value of \(1^2 + 2^2 + 3^2 + ... + n^2\) is NECESSARILY even?
i. n is even
ii. n is a multiple of 6
iii. n is a multiple of 4
(A) only i
(B) only iii
(C) Both i and ii
(D) Both i and iii
(E) Both ii and iii
i. n is even
ii. n is a multiple of 6
iii. n is a multiple of 4
(A) only i
(B) only iii
(C) Both i and ii
(D) Both i and iii
(E) Both ii and iii
04 Mar 2020, 23:14
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contactakanksha
For which of the following values of n , the value of \(1^2 + 2^2 + 3^2 + ... + n^2\) is NECESSARILY even?
i. n is even
ii. n is a multiple of 6
iii. n is a multiple of 4
(A) only i
(B) only iii
(C) Both i and ii
(D) Both i and iii
(E) Both ii and iii
i. n is even
ii. n is a multiple of 6
iii. n is a multiple of 4
(A) only i
(B) only iii
(C) Both i and ii
(D) Both i and iii
(E) Both ii and iii
the value of \(1^2 + 2^2 + 3^2 + ... + n^2\) as even will be same as for \(1+2+3+..+n\)
Let us work out a pattern
1) 1....Odd
2) 1+2... O+E=Odd
3) 1+2+3...O+E+O=Even
4) 1+2+3+4....O+E+O+E=Even
5) 1+2+3+4+5....O+E+O+E+O=Odd
6) 1+2+3+4+5+6....O+E+O+E+O+E=Odd
and so on
So pattern is O,O,E,E,O,O,E,E....
i. n is even....When n is 2 or 6..No
ii. n is a multiple of 6...When n is 6..No or rather whenever n is a multiple of 6, the value will be ODD
iii. n is a multiple of 4...As can be seen from the pattern, all multiples of 4 will give value as EVEN. Also Multiple of 4 means TWO odd and TWO even, so both will add up to EVEN
B
04 Mar 2020, 23:49
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Method 1: Check the partern:
Note that we need an even number of odd terms to add up to an even:
1st term is odd
2nd term is even
3rd term is odd --- 2 odds till here, so even
4th term is even --- 2 odds till here, so even
5th term is odd
6th term is even
7th term is odd --- 4 odds till here, so even
8th term is even --- 4 odds till here, so even
Observe that whenever the number of terms i.e. n is of the form 4k (multiple of 4) or 4k-1, the sum is even
Hence, option B
Method 2: Use summation formula:
The above summation is n(n+1)(2n+1)/6
Clearly:
# if n = 4k, though 2 cancels from the 6 in the denominator, the numerator has a 2 remaining => hence the result is even
# if n = 4k-1, the term (n+1) becomes 4k and hence, from the above logic, the result is even
Hence Option B
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Note that we need an even number of odd terms to add up to an even:
1st term is odd
2nd term is even
3rd term is odd --- 2 odds till here, so even
4th term is even --- 2 odds till here, so even
5th term is odd
6th term is even
7th term is odd --- 4 odds till here, so even
8th term is even --- 4 odds till here, so even
Observe that whenever the number of terms i.e. n is of the form 4k (multiple of 4) or 4k-1, the sum is even
Hence, option B
Method 2: Use summation formula:
The above summation is n(n+1)(2n+1)/6
Clearly:
# if n = 4k, though 2 cancels from the 6 in the denominator, the numerator has a 2 remaining => hence the result is even
# if n = 4k-1, the term (n+1) becomes 4k and hence, from the above logic, the result is even
Hence Option B
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