Math Snap

STEP 1

Assumptions1. The given formula for the sum of squares of the first n natural numbers is $1^{}+^{}+3^{}+\cdots+n^{}=\frac{1}{6} n(n+1)( n+1)$. We need to find the sum of squares of even numbers from to100.

STEP 2

We notice that the sequence $2^{2},4^{2},6^{2},8^{2}, \cdots,98^{2},100^{2}$ is a sequence of squares of the first50 even numbers. We can express each term as $(2k)^{2}$ where $k$ is a natural number from1 to50.

STEP 3

We can rewrite the sequence as $\times1^{2}, \times2^{2}, \times3^{2}, \times^{2}, \cdots, \times49^{2}, \times50^{2}$. This is equivalent to $\times (1^{2}+2^{2}+3^{2}+\cdots+49^{2}+50^{2})$.

STEP 4

We can use the given formula for the sum of squares of the first n natural numbers to calculate the sum in the brackets.
$$1^{2}+2^{2}+3^{2}+\cdots+50^{2}=\frac{1}{6} \times50 \times (50+1) \times (2 \times50+1)$$

STEP 5

Now, calculate the sum in the brackets.
$$1^{2}+2^{2}+3^{2}+\cdots+50^{2}=\frac{1}{} \times50 \times51 \times101 =43,883$$

STEP 6

Now, we can find the sum of squares of even numbers from2 to100 by multiplying the sum we found in the brackets by4.
$$2^{2}+4^{2}+6^{2}+8^{2}+\cdots+98^{2}+100^{2} =4 \times43,883$$

Calculate the sum of squares of even numbers from2 to100.
$$2^{2}+4^{2}+6^{2}+^{2}+\cdots+98^{2}+100^{2} =4 \times43,883 =175,532$$The value of $2^{2}+4^{2}+6^{2}+^{2}+\cdots+98^{2}+100^{2}$ is175,532.