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1^2+2^2+.......+10^2/1+2+.......+10=? A. 7/2 B. 7/3 C. 5 D. 7 E. 14

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Kushchokhani
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1^2+2^2+.......+10^2/1+2+.......+10=? A. 7/2 B. 7/3 C. 5 D. 7 E. 14 [#permalink] New post  18 Sep 2022, 10:59
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\(\frac{1^2+2^2+.......+10^2}{1+2+.......+10}=?\)

(A) 7/2
(B) 7/3
(C) 5
(D) 7
(E) 14
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wickedvikram
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Re: 1^2+2^2+.......+10^2/1+2+.......+10=? A. 7/2 B. 7/3 C. 5 D. 7 E. 14 [#permalink] New post  18 Sep 2022, 11:11
D -7

(n.n+1.2n+1/6 )/(n.n+1/2)
=2n+1/3= 7 for n=10

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Re: 1^2+2^2+.......+10^2/1+2+.......+10=? A. 7/2 B. 7/3 C. 5 D. 7 E. 14 [#permalink] New post  19 Sep 2022, 00:51
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Kushchokhani wrote:
\(\frac{1^2+2^2+.......+10^2}{1+2+.......+10}=?\)

(A) 7/2
(B) 7/3
(C) 5
(D) 7
(E) 14

Solution:

  • We know
    • \(1^2+2^2+3^2....+n^2=\frac{n(n+1)(2n+1)}{6}\) and
    • \(1+2+3+....+n=\frac{n(n+1)}{2}\)
  • Therefore, the value of \(\frac{1^2+2^2+.......+10^2}{1+2+.......+10}=\frac{\frac{10.(10+1).(2.10+1)}{6}}{\frac{10.11}{2}}=\frac{10\times 11\times 21\times 2}{10\times 11\times 6}=7\)


Hence the right answer is Option D
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1^2+2^2+.......+10^2/1+2+.......+10=? A. 7/2 B. 7/3 C. 5 D. 7 E. 14 [#permalink] New post  19 Sep 2022, 08:53
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Kushchokhani wrote:
\(\frac{1^2+2^2+.......+10^2}{1+2+.......+10}=?\)

(A) 7/2
(B) 7/3
(C) 5
(D) 7
(E) 14


I had to look up for the formula!

Sum of squares of n natural numbers = n (n+1) (2n+1)/6

Sum of n natural numbers = n(n+1)/2

If we know these, question becomes simple work.

=\(\frac{1^2+2^2+.......+10^2}{1+2+.......+10}=?\)
= ( 10 (11) (21)/6 ) * ( 10 (11)/2)
= 21/3
= 7

Answer (D)


Bonus:
Sum of squares of first n even numbers = 2n (n+1) (2n+1)/3

Sum of squares of first n odd numbers = n (2n+1) (2n-1)/3
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1^2+2^2+.......+10^2/1+2+.......+10=? A. 7/2 B. 7/3 C. 5 D. 7 E. 14 [#permalink]
19 Sep 2022, 08:53
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