MCQ
$\sum\limits_{m = 1}^n {{m^2}} $ is equal to
  • A
    $\frac{{m(m + 1)}}{2}$
  • B
    $\frac{{m(m + 1)(2m + 1)}}{6}$
  • $\frac{{n(n + 1)(2n + 1)}}{6}$
  • D
    $\frac{{n(n + 1)}}{2}$

Answer

Correct option: C.
$\frac{{n(n + 1)(2n + 1)}}{6}$
c
(c) It is nothing but $\sum n^2$ $= \frac{{n(n + 1)(2n + 1)}}{{6}}$

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Similar questions

$2.\mathop {357}\limits^{ \bullet \,\, \bullet \,\, \bullet } = $
If $\sin \,\theta  + \sqrt 3 \cos \,\theta  = 6x - {x^2} - 11,x \in R$ , $0 \le \theta  \le 2\pi $ , then the equation has solution for
Let $A B C D$ be a square and $E$ be a point outside $A B C D$ such that $E, A, C$ are collinear in that order. Suppose $E B=E D=\sqrt{130}$ and the areas to $\triangle E A B$ and square $A B C D$ are equal. Then, the area of square $A B C D$ is
If the straight line, $2x -3y + 17 = 0$ is perpendicular to the line passing through the points $(7, 17)$ and $(15, \beta )$, then $\beta $ equals:
A number is chosen from first $100$ natural numbers. The probability that the number is even or divisible by $5$, is
if $x_1, x_2, x_3, x_4, x_5$ are five consecutive odd numbers, then their average is:
Ten trucks, numbered $1$ to $10$ , are carrying packets of sugar. Each packet weights either $999\,g$ or $1000\,g$ and each truck carries only the packets equal weights. The combined weight of $1$ packet selected from the first truck,$2$ packets from the second,$4$ packets from the third, and so on, and $2^9$ packet from the tenth truck is $1022870\,g$. The trucks that have the lighter bags are
Given $a^2 + 2a + cosec^2$ $\left( {\frac{\pi }{2}(a + x)} \right)$ $= 0$ then, which of the following holds good?
The sum of $n$ terms of the following series $1 + (1 + x) + (1 + x + {x^2}) + ..........$ will be
Let $A = \{a, b\}, B = \{a, b, c\}.$ What is $\text{A }\cup\text{ B }?$