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📐 Mathematics  ·  Sequences and Series  ·  JEE

Sum of first n squares: 1² + 2² + 3² + ... + n² =

Answer: n(n+1)(2n+1)/6.

  • A n(n+1)/2
  • B n(n+1)(n+2)/6
  • C n(n+1)(2n+1)/6
  • D n²(n+1)/2

Correct answer: C. n(n+1)(2n+1)/6

Explanation: Sum of squares = n(n+1)(2n+1)/6. For n=3: 3×4×7/6 = 14.

Each bar grows by the common difference d=3a1=2a2=5a3=8a4=11d=3

Bar heights for AP 2, 5, 8, 11 with the common difference d=3 marked between consecutive terms.

Concept context

Arithmetic progressions, geometric progressions, and sums

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