C-13-Q076mediumsingle_mcq

The sum of the squares of the first $n$ natural odd numbers is —

a $\dfrac{n(2n - 1)(2n + 1)}{3}$
b $\dfrac{n(n + 1)(2n + 1)}{6}$
c $\dfrac{n^{2}(n + 1)^{2}}{4}$
d $n(n + 1)$

ব্যাখ্যা

The sum of the squares of the first $n$ odd numbers is the standard identity $\frac{n(2n-1)(2n+1)}{3}$ . For example, with $n=2$ it gives $1^2+3^2=\frac{2\times3\times5}{3}=10$ , confirming the formula.

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Finite Series — সব প্রশ্ন দেখুন (160টি) →

The sum of the squares of the first $n$ natural odd numbers is —

\dfrac{n(2n - 1)(2n + 1)}{3}

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

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

n(n + 1)

The sum of the squares of the first nnn natural odd numbers is —

ব্যাখ্যা

এই অধ্যায়ের আরও প্রশ্ন

The sum of the squares of the first nnn natural odd numbers is —

ব্যাখ্যা

এই অধ্যায়ের আরও প্রশ্ন

The sum of the squares of the first $n$ natural odd numbers is —

The sum of the squares of the first $n$ natural odd numbers is —