# Observe the Following Pattern 1 = 1 2 { 1 × ( 1 + 1 ) } 1 + 2 = 1 2 { 2 × ( 2 + 1 ) } 1 + 2 + 3 = 1 2 { 3 × ( 3 + 1 ) } 1 + 2 + 3 + 4 = 1 2 { 4 × ( 4 + 1 ) } - Mathematics

Observe the following pattern $1 = \frac{1}{2}\left\{ 1 \times \left( 1 + 1 \right) \right\}$
$1 + 2 = \frac{1}{2}\left\{ 2 \times \left( 2 + 1 \right) \right\}$
$1 + 2 + 3 = \frac{1}{2}\left\{ 3 \times \left( 3 + 1 \right) \right\}$
$1 + 2 + 3 + 4 = \frac{1}{2}\left\{ 4 \times \left( 4 + 1 \right) \right\}$and find the values of following:

31 + 32 + ... + 50

#### Solution

Observing the three numbers for right hand side of the equalities:
The first equality, whose biggest number on the LHS is 1, has 1, 1 and 1 as the three numbers.
The second equality, whose biggest number on the LHS is 2, has 2, 2 and 1 as the three numbers.
The third equality, whose biggest number on the LHS is 3, has 3, 3 and 1 as the three numbers.
The fourth equality, whose biggest number on the LHS is 4, has 4, 4 and 1 as the three numbers.
Hence, if the biggest number on the LHS is n, the three numbers on the RHS will be nnand 1.
Using this property, we can calculate the sums for (i) and (ii) as follows:

The sum can be expressed as the difference of the two sums as follows:

$31 + 32 + . . . . . + 50 = (1 + 2 + 3 + . . . . . . + 50) - ( 1 + 2 + 3 + . . . . . . + 30)$

The result of the first bracket is exactly the same as in part (i).

$1 + 2 + . . . . + 50 = 1275$

Then, the second bracket:

$1 + 2 + . . . . . . + 30 = \frac{1}{2}\left( 30 \times \left( 30 + 1 \right) \right)$

Finally, we have:

$31 + 32 + . . . . + 50 = 1275 - 465 = 810$

Is there an error in this question or solution?

#### APPEARS IN

RD Sharma Class 8 Maths
Chapter 3 Squares and Square Roots
Exercise 3.2 | Q 9.2 | Page 19