C-13-Q066mediumsingle_mcq

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

a $\dfrac{n(n+1)(2n+1)}{6}$
b $\dfrac{n^{2}(n+1)}{2}$
c $\left[\dfrac{n(n+1)}{2}\right]^{2}$
d $\dfrac{n^{2}(n-1)^{2}}{4}$

ব্যাখ্যা

The sum of the cubes of the first $n$ natural numbers equals $\left[\frac{n(n+1)}{2}\right]^2$ , which is the square of the sum of the first $n$ natural numbers. This is the well-known cube-sum identity.

এই অধ্যায়ের আরও প্রশ্ন

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 —
C-13-Q076The sum of the squares of the first natural odd numbers is —

Finite Series — সব প্রশ্ন দেখুন (160টি) →

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

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

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

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

\dfrac{n^{2}(n-1)^{2}}{4}

13+23+33+⋯+n3=?1^{3} + 2^{3} + 3^{3} + \cdots + n^{3} = ?13+23+33+⋯+n3=?

ব্যাখ্যা

এই অধ্যায়ের আরও প্রশ্ন

13+23+33+⋯+n3=?1^{3} + 2^{3} + 3^{3} + \cdots + n^{3} = ?13+23+33+⋯+n3=?

ব্যাখ্যা

এই অধ্যায়ের আরও প্রশ্ন

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

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