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36^2+37^2+38^2+39^2+40^2+41^2+42^2+43^2+44^2=
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06 Sep 2021, 09:43
06 Sep 2021, 09:43
Expert Reply
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Question Stats:
60% (03:12) correct
40% (02:23) wrong based on 10 sessions
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\(36^2\) + \(37^2\) + \(38^2\) + \(39^2\) + \(40^2\) + \(41^2\) + \(42^2\)+ \(43^2\) + \(44^2\) =
(A) 14400
(B) 14440
(C) 14460
(D) 14500
(E) 14520
(A) 14400
(B) 14440
(C) 14460
(D) 14500
(E) 14520
Re: 36^2+37^2+38^2+39^2+40^2+41^2+42^2+43^2+44^2=
[#permalink]
07 Sep 2021, 06:22
07 Sep 2021, 06:22
2
Answer is (C) 14460
Try 1: just square the last digits of all numbers like 6^2 + 7^2 + ... 4^2 and it was 260 , so only choice ending with 60 is C.
Try 2: Sum of squares from 1 to n is n*(n+1)(2n+1)/6
so use the formula to sum up to 44^2 and subtract sum up to 35^2 and it comes to 14460.
Try 1: just square the last digits of all numbers like 6^2 + 7^2 + ... 4^2 and it was 260 , so only choice ending with 60 is C.
Try 2: Sum of squares from 1 to n is n*(n+1)(2n+1)/6
so use the formula to sum up to 44^2 and subtract sum up to 35^2 and it comes to 14460.
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36^2+37^2+38^2+39^2+40^2+41^2+42^2+43^2+44^2=
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21 Sep 2024, 21:40
21 Sep 2024, 21:40
1
We need to find the sum of \(36^2+37^2+38^2+39^2+40^2+41^2+42^2+43^2+44^2\)
We know that sum of first n positive integers starting from 1 = \(\frac{n*(n+1)*(2n+1)}{6}\)
\(36^2+37^2+38^2+39^2+40^2+41^2+42^2+43^2+44^2\) = \(1^2 + 2^2 + .... + 36^2+37^2+38^2+39^2+40^2+41^2+42^2+43^2+44^2\) - \(1^2 + 2^2 + .... + 34^2+35^2\)
= Sum of Squares from 1 to 44 - Sum of Squares from 1 to 35 = \(\frac{44*(44+1)*(2*44+1)}{6}\) - \(\frac{35*(35+1)*(2*35+1)}{6}\)
= \(\frac{44*45*89}{6}\) - \(\frac{35*36*71}{6}\)
= 29,370 - 14,910
= 14,460
So, Answer will C
Hope it helps!
Watch the following video to MASTER Sequence problems
We know that sum of first n positive integers starting from 1 = \(\frac{n*(n+1)*(2n+1)}{6}\)
\(36^2+37^2+38^2+39^2+40^2+41^2+42^2+43^2+44^2\) = \(1^2 + 2^2 + .... + 36^2+37^2+38^2+39^2+40^2+41^2+42^2+43^2+44^2\) - \(1^2 + 2^2 + .... + 34^2+35^2\)
= Sum of Squares from 1 to 44 - Sum of Squares from 1 to 35 = \(\frac{44*(44+1)*(2*44+1)}{6}\) - \(\frac{35*(35+1)*(2*35+1)}{6}\)
= \(\frac{44*45*89}{6}\) - \(\frac{35*36*71}{6}\)
= 29,370 - 14,910
= 14,460
So, Answer will C
Hope it helps!
Watch the following video to MASTER Sequence problems
