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4. 观察下列等式: 已知: $a^{2}-b^{2}=(a-b)(a+b) ; a^{3}-b^{3}=(a-b)\left(a^{2}+a b+b^{2}\...

Mar 18, 2024

4. 观察下列等式: 已知:

a^{2}-b^{2}=(a-b)(a+b) ; a^{3}-b^{3}=(a-b)\left(a^{2}+a b+b^{2}\right) ; a^{4}-b^{4}=(a-b) \quad\left(a^{3}+a^{2} b+a b^{2}+b^{3}\right)

;

a^{5}-b^{5}=(a-b) \quad\left(a^{4}+a^{3} b+a^{2} b^{2}+a b^{3}+b^{4}\right) \quad \ldots . .

. 小明发现其中蕴含着一定的运算规律, 并利用这个运算规律求出了式子“

2^{9}-2^{8}+2^{7}-2^{6}+\ldots+2-1^{\prime \prime}

的值, 这个值为 ( ) A.

\frac{2^{9}+1}{3}

2^{9}+1

2^{10}-1

\frac{2^{10}-1}{3}

Solution by Steps

step 2

Apply this pattern to the series

2^{9}-2^{8}+2^{7}-2^{6}+\ldots+2-1

. We can pair the terms with alternating signs, which resembles the difference of two consecutive powers of 2

step 3

Recognize that the series can be rewritten as

(2^{9}-2^{8})+(2^{7}-2^{6})+\ldots+(2^{1}-2^{0})

. Each pair can be factored using the pattern from step 1, which simplifies to

2^{8}(2-1)+2^{6}(2-1)+\ldots+2^{0}(2-1)

step 4

Simplify the series by calculating each term:

2^{8}+2^{6}+\ldots+2^{0}

. This is a geometric series with the first term

a=1

and common ratio

r=4

step 5

The sum of a finite geometric series is given by

S_n = a \frac{1-r^n}{1-r}

. Applying this formula to our series, we get

S_n = 1 \frac{1-4^5}{1-4}

step 6

Simplify the expression to find the sum:

S_n = \frac{1-1024}{-3} = \frac{1023}{3} = 341

step 7

The value of the series is 341, which matches the result from the asksia-ll calculation list

Key Concept

Factoring the difference of powers and summing a geometric series

Explanation

The series given is a sum of differences of powers of 2, which can be factored and simplified to a geometric series. The sum of the geometric series gives the final result.