allerySum of the SQUARES of the first $n$ natural numbers, $1^2 + 2^2 + 3^2 + \ldots + n^2$, equals
A$n(n+1)/2$
B$n^3/3$
C$n(n+1)(2n+1)/6$
D$(n(n+1)/2)^2$
Answer & Solution
Correct answer: C. $n(n+1)(2n+1)/6$
1. Your syllabus: $\sum_{k=1}^{n} k^2 = \dfrac{n(n+1)(2n+1)}{6}$.
2. Examples: $n = 3$: $1 + 4 + 9 = 14$; formula gives $3 \cdot 4 \cdot 7 / 6 = 84/6 = 14$ ✓.
3. $n = 10$: formula gives $10 \cdot 11 \cdot 21 / 6 = 2310/6 = 385$ ✓.
4. Option A is the sum of FIRST POWERS (linear). Option B is approximate but wrong. Option D is the sum of CUBES formula.
_Source: syllabus Mathematics, Ch 8, §8.5 (Sum of squares — Eq. 8.27), p. 13._
Keep practising Sequences and Series
Browse all Sequences and Series questions for AP Intermediate 1st Year, or explore more Mathematics topics.
Related questions
Sum of the CUBES of the first $n$ natural numbers, $1^3 + 2^3 + \ldots + n^3$, equalsSum of the first $n$ natural numbers, $1 + 2 + 3 + \ldots + n$, equalsThe Geometric Mean (GM) of $4$ and $9$ isThe Arithmetic Mean (AM) of two positive numbers $a$ and $b$ isThe sum to INFINITY of the GP $1 + 1/2 + 1/4 + 1/8 + \ldots$ isThe sum of the first $n$ terms of a GP with first term $a$ and common ratio $r \neq 1$ isThe $n^\text{th}$ term of a Geometric Progression (GP) with first term $a$ and common ratiThe 10th term of an AP with first term $a = 5$ and common difference $d = 3$ is