Find the sum of squares of first 15 natural numbers
Answers
If you know the sum of squares formula (something I've never seen tested in a real GMAT question), you can use it to sum the first 30 squares, then subtract 1240 to get the answer. The sum of the squares of the first n positive integers is given by (n)(n+1)(2n+1)/6, though I'd be very surprised to find a real GMAT question where this was at all helpful to know.
Or, if you're familiar with remainder/modular arithmetic (if not, this solution might seem confusing), you might notice that when you add any three consecutive perfect squares, the remainder will be 2 when you divide by 3 (since one of the squares will be divisible by 3, and the other two will both give a remainder of 1). So here we're adding five sets of three numbers, where each set gives a remainder of 2 when divided by 3. Since we can add remainders, the remainder when we divide 16^2 + 17^2 + ... + 30^2 by 3 will be the remainder when we divide 5*2 = 10 by 3, so the remainder will be 1.
You could also estimate here. Notice that 15^2 + 16^2 + 17^2 + 18^2 + 19^2 is quite a bit bigger than 15^2 + 15^2 + ... + 15^2 = 5*15^2, and quite a bit smaller than 20^2 + 20^2 + ... + 20^2. Similarly, the next block of five squares add to something quite a bit bigger than 5*20^2, and quite a bit smaller than 5*25^2, and similarly for the last block. So our sum S must satisfy the following inequality, and shouldn't be anywhere close to the extreme low or high values:
5*15^2 + 5*20^2 + 5^25^2 < S < 5*20^2 + 5*25^2 + 5*30^2
5(15^2 + 20^2 + 25^2) < S < 5(20^2 + 25^2 + 30^2)
5(1250) < S < 5(1925)
6250 < S < 9625
Or you can proceed algebraically, using the (x +y)^2 = x^2 + 2xy + y^2 pattern to relate the sum to the one given in the question:
16^2 + 17^2 + 18^2 + ... + 29^2 + 30^2 =
(1 + 15)^2 + (2 + 15)^2 + (3 + 15)^2 + ... + (14 + 15)^2 + (15 + 15)^2 =
(1^2 + 2*15 + 15^2) + (2^2 + 2*2*15 + 15^2) + (3^2 + 2*3*15 + 15^2) + ... + (14^2 + 2*14*15 + 15^2) + (15^2 + 2*15*15 + 15^2) =
(1^2 + 2^2 + ... + 15^2) + 2*15(1 + 2 + 3 + ... + 14 + 15) + 15*15^2
= 1240 + 3600 + 3375
= 8215
Solution:
Sum of squares of first n natural numbers = n (n+1) (2n+1) / 6
So, in this question n is 15 and by following this we will put the value of n as 15 in the formula,
= n (n+1) (2n+1) / 6
= 15 (15+1) (2 x 15 + 1) / 6
= 15 x 16 x 31 / 6
= 15 x 496 / 6
= 1240
I hope that this solution helps you
if yes then please mark me as the brainliest
Answer = 1240