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Observe the Following Pattern 1 2 = 1 6 [ 1 × ( 1 + 1 ) × ( 2 × 1 + 1 ) ] 1 2 + 2 2 = 1 6 [ 2 × ( 2 + 1 ) × ( 2 × 2 + 1 ) ] 1 2 + 2 2 + 3 2 = 1 6 [ 3 × ( 3 + 1 ) × ( 2 × 3 + 1 ) ] - Mathematics

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प्रश्न

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 + 4+ ... + 102

 

 

उत्तर

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 nn, 1, 2, n and 1.
Using this property, we can calculate the sums for (i) and (ii) as follows: 

\[ 1^2 + 2^2 + . . . . . . . + {10}^2 = \frac{1}{6} \times \left[ 10 \times \left( 10 + 1 \right) \times \left( 2 \times 10 + 1 \right) \right]\]
\[= \frac{1}{6} \times \left[ 10 \times 11 \times 12 \right] \]
\[ = 385\]

 

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अध्याय 3: Squares and Square Roots - Exercise 3.2 [पृष्ठ २०]

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आरडी शर्मा Mathematics [English] Class 8
अध्याय 3 Squares and Square Roots
Exercise 3.2 | Q 10.1 | पृष्ठ २०

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