kalita wrote:
What is the average of the following numbers?
1^2, 2^2, 3^2...10^2.
A. 5.5
B. 25
C. 38.5
D. 55
E. 385
Easy problem but may take time. The book explains it the following way which I am wondering about:
“∑_(n=1)^10▒〖n^2=〗 1/6 n(n+1)(2n+1)=385. The average is therefore 385/10 = 38.5”I do not understand the bits and pieces of the above. Moreover, I am wondering if there is an alternative shortcut to solve this type of problem?
Or, for example, is there a way to calculate the sum of 1^2 through…20^2 using the above shortcut or another method?
What is the number property tested here?
Thanks in advance.
Yes, even if you do not know the formula, a bit of logic and number properties can help you solve the question. The numbers will be
1, 4, 9, 16, 25, 36, 49, 64, 81, 100
Note that as you go to higher numbers, the square spreads out more. the difference between 8^2 and 9^2 will be much more than the difference between 6^2 and 7^2.
Only 3 numbers are greater than 50 and that too not all very far from it but we do have a lot of very small numbers (1, 4, 9) so the average is certainly less than 50.
In increasing order, 30 lies about mid way (between 25 and 36) but deviation of the numbers to the right of it is much more than the deviation to the left of it. Hence, the average should certainly be greater than 25.
Answer (C)
Great alternative approach. Very insightful. Thank you.