C-13-Q065mediumsingle_mcq

$1^{2} + 2^{2} + 3^{2} + \cdots + n^{2} = ?$

a $\dfrac{n(n+1)}{2}$
b $\dfrac{n(n+1)(2n+1)}{6}$
c $\left[\dfrac{n(n+1)}{2}\right]^{2}$
d $\dfrac{n(n+1)(2n-1)}{6}$

ব্যাখ্যা

The sum of the squares of the first $n$ natural numbers is the established formula $\frac{n(n+1)(2n+1)}{6}$ . It is a standard identity that must be memorized for this chapter.

এই অধ্যায়ের আরও প্রশ্ন

C-13-Q066
C-13-Q067
C-13-Q068
C-13-Q069
C-13-Q070
C-13-Q071
C-13-Q072If , the value of is —
C-13-Q073If the sum of the cubes of the first natural numbers is , the value of is —
C-13-Q074If the sum of the cubes of the first natural numbers is , the value of is —
C-13-Q075The sum of the first natural even numbers is —

Finite Series — সব প্রশ্ন দেখুন (160টি) →

$1^{2} + 2^{2} + 3^{2} + \cdots + n^{2} = ?$

\dfrac{n(n+1)}{2}

\dfrac{n(n+1)(2n+1)}{6}

\left[\dfrac{n(n+1)}{2}\right]^{2}

\dfrac{n(n+1)(2n-1)}{6}

ব্যাখ্যা

The sum of the squares of the first $n$ natural numbers is the established formula $\frac{n(n+1)(2n+1)}{6}$ . It is a standard identity that must be memorized for this chapter.

12+22+32+⋯+n2=?1^{2} + 2^{2} + 3^{2} + \cdots + n^{2} = ?12+22+32+⋯+n2=?

ব্যাখ্যা

এই অধ্যায়ের আরও প্রশ্ন

12+22+32+⋯+n2=?1^{2} + 2^{2} + 3^{2} + \cdots + n^{2} = ?12+22+32+⋯+n2=?

ব্যাখ্যা

এই অধ্যায়ের আরও প্রশ্ন

$1^{2} + 2^{2} + 3^{2} + \cdots + n^{2} = ?$

$1^{2} + 2^{2} + 3^{2} + \cdots + n^{2} = ?$