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Question
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:
1 + 2 + 3 + 4 + 5 + ... + 50
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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 n, nand 1.
Using this property, we can calculate the sums for (i) and (ii) as follows:
\[(i) 1 + 2 + 3 + . . . . . . . . + 50 = \frac{1}{2} \times 50 \times (50 + 1) = 1275\]
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Observe the following pattern \[1^2 = \frac{1}{6}\left[ 1 \times \left( 1 + 1 \right) \times \left( 2 \times 1 + 1 \right) \right]\]
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\[ 1^2 + 2^2 + 3^2 = \frac{1}{6}\left[ 3 \times \left( 3 + 1 \right) \times \left( 2 \times 3 + 1 \right) \right]\]
\[ 1^2 + 2^2 + 3^2 + 4^2 = \frac{1}{6}\left[ 4 \times \left( 4 + 1 \right) \times \left( 2 \times 4 + 1 \right) \right]\] and find the values :
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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 :
52 + 62 + 72 + 82 + 92 + 102 + 112 + 122
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