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mba_aspirant911: Junior | Next Rank: 30 Posts; Posts: 13; Joined: Sun Oct 25, 2009 6:30 pm; Thanked: 1 times

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by mba_aspirant911 » Sat Dec 04, 2010 1:56 am

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

Need some help in the approach for this one. I cannot see any obvious sequence/series pattern.

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Source: Beat The GMAT — Problem Solving | Sat Dec 04, 2010 1:56 am

mba_aspirant911: Junior | Next Rank: 30 Posts; Posts: 13; Joined: Sun Oct 25, 2009 6:30 pm; Thanked: 1 times

by mba_aspirant911 » Sat Dec 04, 2010 2:09 am

Actually, just realised it is simplified if 40 is taken as a baseline:

36 = (40 - 4) ^ 2
37 = (40 - 3) ^ 2
....
44 = (40 + 4) ^ 2

36^2 = 40^2 - 320 + 16
37^2 = 40^2 - 240 + 9
....
44^2 = 40^2 + 320 + 14

Therefore there are 9*40^2 = 1600 * 9
Sum of 16 + 9 + 4 + 1 = 30

30 * 2 (due to both sides of 40) = 60

= 14400 + 60 = 14460

C

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junegmat221: Junior | Next Rank: 30 Posts; Posts: 24; Joined: Wed Apr 28, 2010 4:03 am

by junegmat221 » Sat Dec 04, 2010 3:38 am

Absolutely great...
This is the most concise solution.
But if u needed a time consuming solution , it would be this.
It would surely take ur 2 mins off.
Use the formula (n(n+1)(2n+1))/6

First compute for first 44 natural numbers..
Second compute for first 35 natural numbers.(In exam time if we didnt remain calm we would go with 36 instead of 35, because of which , this is a dangerous solution).

Subtract both of them.
You will find answer to be C Again