The sum of the squares of the first n positive odd integers is $\frac{(2 n-1)(2 n)(2 n+1)}{6}$
Use this expression to find the sum $1^{2}+3^{2}+5^{2}+\cdots+75^{2}$
A) 5628
B) 73,150
C) 1425
D) 7,700,625
E) 5776

Question

The sum of the squares of the first  n  positive odd integers is  $\frac{(2 n-1)(2 n)(2 n+1)}{6}$ 
Use this expression to find the sum $1^{2}+3^{2}+5^{2}+\cdots+75^{2}$
A) 5628
B) 73,150
C) 1425
D)  7,700,625 
E) 5776

Quizplus · Accepted Answer

The sum of the squares of the first n positive odd integers is given by the formula $\frac{(2 n-1)(2 n)(2 n+1)}{6}$. To find the sum $1^{2}+3^{2}+5^{2}+\cdots+75^{2}$, we need to substitute n = 38 into the formula.$\frac{(2 \times 38-1)(2 \times 38)(2 \times 38+1)}{6} = \frac{(75)(76)(77)}{6} = 73,150$

The Sum of the Squares of the First N (2n−1)(2n)(2n+1)6\frac{(2 n-1)(2 n)(2 n+1)}{6}6(2n−1)(2n)(2n+1)​

The Sum of the Squares of the First N $\frac{(2 n-1)(2 n)(2 n+1)}{6}$