Observe the following pattern 12-161 - 1-1 - 2 - 1-1 12-22-162 - 2-1 - 2 - 2-1 12-22-32-163 - 3-1 - 2 - 3-112-22-32-42-164 - 4-1 - 2 - 4-1and find the values of each of the following-i 12-22-32-42-102ii 52-62-72-82-92-102-112-122

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Writing Number Patterns and Rules Related to Them

Observe the f...

Question

Observe the following pattern
$1^{2} = \frac{1}{6} [1 \times (1 + 1) \times (2 \times 1 + 1)] 1^{2} + 2^{2} = \frac{1}{6} [2 \times (2 + 1) \times (2 \times 2 + 1)] 1^{2} + 2^{2} + 3^{2} = \frac{1}{6} [3 \times (3 + 1) \times (2 \times 3 + 1)] 1^{2} + 2^{2} + 3^{2} + 4^{2} = \frac{1}{6} [4 \times (4 + 1) \times (2 \times 4 + 1)]$
and find the values of each of the following:
(i) 1² + 2² + 3² + 4²+ ... + 10²
(ii) 5² + 6² + 7² + 8² + 9² + 10² + 11² + 12²

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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:

$(i) 1^{2} + 2^{2} + . . . . . . . + 10^{2} = \frac{1}{6} \times [10 \times (10 + 1) \times (2 \times 10 + 1)]$

$= \frac{1}{6} \times [10 \times 11 \times 12] = 385$ .

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

$5^{2} + 6^{2} + . . . . . . . + 12^{2} = (1^{2} + 2^{2} + . . . . . . + 12^{2}) - (1^{2} + 2^{2} + . . . . . . + 4^{2})$

The sum of the first bracket on the RHS:

$1^{2} + 2^{2} + . . . . . + 12^{2} = \frac{1}{6} [12 \times (12 + 1) \times (2 \times 12 + 1)] = 650$

The second bracket is:

$1^{2} + 2^{2} + . . . . . . + 4^{2} = \frac{1}{6} \times [4 \times (4 + 1) \times (2 \times 4 + 1)] = \frac{1}{6} \times 4 \times 5 \times 9 = 30$

Finally, the wanted sum is:

$5^{2} + 6^{2} + . . . . . . + 12^{2} = (1^{2} + 2^{2} + . . . . . + 12^{2}) - (1^{2} + 2^{2} + . . . . . + 12^{2}) = 650 - 30 = 620$

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Similar questions

Using the given pattern, find the missing numbers.
$1^{2} + 2^{2} + 2^{2} = 3^{2}$
$2^{2} + 3^{2} + 6^{2} = 7^{2}$
$3^{2} + 4^{2} + 12^{2} = 13^{2}$
$4^{2} + 5^{2} + . . .^{2} = 21^{2}$
$5^{2} + . . .^{2} + 30^{2} = 31^{2}$
$6^{2} + 7^{2} + . . . .^{2} = . . . .^{2}$ .

Q. Using the given pattern, find the missing numbers:

1^{2} + 2^{2} + 2^{2} = 3^{2}

2^{2} + 3^{2} + 6^{2} = 7^{2}

3^{2} + 4^{2} + 12^{2} = 13^{2}

4^{2} + 5^{2} +

___

= 21^{2}

5^{2} +

___

+ 30^{2} = 31^{2}

6^{2} + 7^{2} +

___

=

____

$1 + \frac{1^{2} + 2^{2}}{2!} + \frac{1^{2} + 2^{2} + 3^{2}}{3!} + \frac{1^{2} + 2^{2} + 3^{2} + 4^{2}}{4!} + . . . \infty = \frac{b}{6} e$ . Find $b$

Q. If

\frac{1 \times 2^{2} + 2 \times 3^{2} + 3 \times 4^{2} + \dots + n (n + 1)^{2}}{1^{2} \times 2 + 2^{2} \times 3 + 3^{2} \times 4 + \dots + n^{2} (n + 1)} = \frac{8}{7}

, then

n =

1 + 2 + 3 + 4 + . . . . . + n = ?

1 + 2 + 3 + 4 + . . . . . + (n - 1) = ?

1 + 1 + 1 + 1 + . . . . . + n = ?

1 + 1 + 1 + 1 + . . . . . . + (n - 1) = ?

1^{2} + 2^{2} + 3^{2} + 4^{2} + . . . . + n^{2} = ?

1^{2} + 2^{2} + 3^{2} + 4^{2} + . . . . + {(n - 1)}^{2} = ?

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Observe the following pattern 12=161×1+1×2×1+1 12+22=162×2+1×2×2+1 12+22+32=163×3+1×2×3+112+22+32+42=164×4+1×2×4+1 and find the values of each of the following: (i) 12 + 22 + 32 + 42 + ... + 102 (ii) 52 + 62 + 72 + 82 + 92 + 102 + 112 + 122