Question
Download Solution PDF1 + 4 + 9 + 16 + 25 + ..... + 81 + 100 + 81 + ...... + 16 + 9 + 4 + 1 = ?
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFGiven:
1 + 4 + 9 + 16 + 25 + ..... + 81 + 100 + 81 + ...... + 16 + 9 + 4 + 1
Concept used:
Sum of Square of 1st 'n' natural numbers is given by
12 + 22 + 32 + ..... + n2 = \(\frac{n(n+1)(2n+1)}{6}\)
Calculation:
According to the question,
1 + 4 + 9 + 16 + 25 + ..... + 81 + 100 + 81 + ..... + 16 + 9 + 4 + 1
⇒ 2 (1 + 4 + 9 + 16 + 25 + ..... + 81) + 100
⇒ 2 (12 + 22 + 32 + 42 + .... + 92) + 100
By using the formula for sum of squares of 1st '9' natural numbers,
⇒ 2 \( [\frac{9(9+1)(18 +1)}{6}]\) + 100
⇒ 2 \((\frac{9\; \times \;10 \;\times \;19}{6} )\) + 100
⇒ 570 + 100 = 670
∴ The sum of the given series is 670.
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