If (1 + 3 + 5 + … + p) + (1 + 3 + 5 + … + q)
= (1 + 3 + 5 + … + r) where each set of parentheses contains the sum of consecutive odd integers as shown, the smallest possible value of p + q + r, (where p > 6) is -
Text Solution
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We know that
1 + 3 + 5 + … + (2k – 1) = k 2
Thus, the given equation can be written as
⇒ (p + 1) 2 + (q + 1) 2 = (r + 1) 2 Therefore, (p + 1, q + 1, r + 1) forms a Pythagorean triplet. As p > 6, p + 1 > 7.
The first Pythagorean triplet containing a number > 7 is (6, 8, 10).
∴ We may take p + 1 = 8, q + 1 = 6, r + 1 = 10
⇒ p + q + r = 21.
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