QuestionFind the sum of squares of even numbers: $2^{2}+4^{2}+6^{2}+\cdots+100^{2}$ . Use $1^{2}+2^{2}+\cdots+n^{2}=\frac{1}{6} n(n+1)(2 n+1)$ .

Studdy Solution

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.
$2^{2}+4^{2}+6^{2}+8^{2}+\cdots+98^{2}+100^{2} =4 \times43,883$

STEP 7

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.

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QuestionFind the sum of squares of even numbers: $2^{2}+4^{2}+6^{2}+\cdots+100^{2}$ . Use $1^{2}+2^{2}+\cdots+n^{2}=\frac{1}{6} n(n+1)(2 n+1)$ .

Studdy Solution

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.
$2^{2}+4^{2}+6^{2}+8^{2}+\cdots+98^{2}+100^{2} =4 \times43,883$

STEP 7

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.

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QuestionFind the sum of squares of even numbers: 22+42+62+⋯+10022^{2}+4^{2}+6^{2}+\cdots+100^{2}22+42+62+⋯+1002. Use 12+22+⋯+n2=16n(n+1)(2n+1)1^{2}+2^{2}+\cdots+n^{2}=\frac{1}{6} n(n+1)(2 n+1)12+22+⋯+n2=61​n(n+1)(2n+1).

Studdy Solution

STEP 1

Assumptions1. The given formula for the sum of squares of the first n natural numbers is 1++3+⋯+n=16n(n+1)(n+1)1^{}+^{}+3^{}+\cdots+n^{}=\frac{1}{6} n(n+1)( n+1)1++3+⋯+n=61​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 22,42,62,82,⋯ ,982,10022^{2},4^{2},6^{2},8^{2}, \cdots,98^{2},100^{2}22,42,62,82,⋯,982,1002 is a sequence of squares of the first50 even numbers. We can express each term as (2k)2(2k)^{2}(2k)2 where kkk is a natural number from1 to50.

STEP 3

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.12+22+32+⋯+502=16×50×(50+1)×(2×50+1)1^{2}+2^{2}+3^{2}+\cdots+50^{2}=\frac{1}{6} \times50 \times (50+1) \times (2 \times50+1)12+22+32+⋯+502=61​×50×(50+1)×(2×50+1)

STEP 5

Now, calculate the sum in the brackets.12+22+32+⋯+502=1×50×51×101=43,8831^{2}+2^{2}+3^{2}+\cdots+50^{2}=\frac{1}{} \times50 \times51 \times101 =43,88312+22+32+⋯+502=1​×50×51×101=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 by4.22+42+62+82+⋯+982+1002=4×43,8832^{2}+4^{2}+6^{2}+8^{2}+\cdots+98^{2}+100^{2} =4 \times43,88322+42+62+82+⋯+982+1002=4×43,883

STEP 7

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QuestionFind the sum of squares of even numbers: $2^{2}+4^{2}+6^{2}+\cdots+100^{2}$ . Use $1^{2}+2^{2}+\cdots+n^{2}=\frac{1}{6} n(n+1)(2 n+1)$ .

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.

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.

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

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

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