Advertisements
Advertisements
Question
Observe the following pattern \[1^2 = \frac{1}{6}\left[ 1 \times \left( 1 + 1 \right) \times \left( 2 \times 1 + 1 \right) \right]\]
\[ 1^2 + 2^2 = \frac{1}{6}\left[ 2 \times \left( 2 + 1 \right) \times \left( 2 \times 2 + 1 \right) \right]\]
\[ 1^2 + 2^2 + 3^2 = \frac{1}{6}\left[ 3 \times \left( 3 + 1 \right) \times \left( 2 \times 3 + 1 \right) \right]\]
\[ 1^2 + 2^2 + 3^2 + 4^2 = \frac{1}{6}\left[ 4 \times \left( 4 + 1 \right) \times \left( 2 \times 4 + 1 \right) \right]\] and find the values :
12 + 22 + 32 + 42 + ... + 102
Advertisements
Solution
Observing the six numbers on the RHS of the equalities:
The first equality, whose biggest number on the LHS is 1, has 1, 1, 1, 2, 1 and 1 as the six numbers.
The second equality, whose biggest number on the LHS is 2, has 2, 2, 1, 2, 2 and 1 as the six numbers.
The third equality, whose biggest number on the LHS is 3, has 3, 3, 1, 2, 3 and 1 as the six numbers.
The fourth equality, whose biggest number on the LHS is 4, has numbers 4, 4, 1, 2, 4 and 1 as the six numbers.
Note that the fourth number on the RHS is always 2 and the sixth number is always 1. The remaining numbers are equal to the biggest number on the LHS.
Hence, if the biggest number on the LHS is n, the six numbers on the RHS would be n, n, 1, 2, n and 1.
Using this property, we can calculate the sums for (i) and (ii) as follows:
\[ = 385\]
APPEARS IN
RELATED QUESTIONS
From the pattern, we can say that the sum of the first n positive odd numbers is equal to the square of the n-th positive number. Putting that into formula:
1 + 3 + 5 + 7 + ... n = n2, where the left hand side consists of n terms.
Which of the following number square of even number?
121
Find the squares of the following numbers using diagonal method:
273
Find the squares of the following numbers using diagonal method:
171
Find the square of the following number:
503
Find the square of the following number:
405
The sum of two perfect squares is a perfect square.
All numbers of a pythagorean triplet are odd.
Write two Pythagorean triplets each having one of the numbers as 5.
