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不羈 發問於 教育及參考書教學 · 10 年前

f.2 math

expend (a-b)(a+b)(a^2+b^2)(a^4+b^4)(a^8+b^8)(a^16+b^16).....(a^1024+b^1024)

describe your answer simply

3 個解答

評分
  • 10 年前
    最愛解答

    Hi ! I am lop****** , feel happy to answer your question.

    Q : expend (a-b)(a+b)(a^2+b^2)(a^4+b^4)(a^8+b^8)(a^16+b^16).....(a^1024+b^1024)

    A :

    By th formula (x+y)(x-y) = x^2 - y^2 and (x^n)^2 = x^2n :

    (a-b)(a+b)(a^2+b^2)(a^4+b^4)(a^8+b^8)(a^16+b^16).....(a^1024+b^1024)

    = (a^2-b^2)(a^2+b^2)(a^4+b^4)(a^8+b^8)(a^16+b^16)(a^32+b^32)(a^64+b^64)(a^128+b^128)(a^256+b^256)(a^512+b^512)(a^1024+b^1024)

    = (a^4-b^4)(a^4+b^4)(a^8+b^8)(a^16+b^16)(a^32+b^32)(a^64+b^64)(a^128+b^128)(a^256+b^256)(a^512+b^512)(a^1024+b^1024)

    = (a^8-b^8)(a^8+b^8)(a^16+b^16)(a^32+b^32)(a^64+b^64)(a^128+b^128)(a^256+b^256)(a^512+b^512)(a^1024+b^1024)

    = (a^16-b^16)(a^16+b^16)(a^32+b^32)(a^64+b^64)(a^128+b^128)(a^256+b^256)(a^512+b^512)(a^1024+b^1024)

    = (a^32-b^32)(a^32+b^32)(a^64+b^64)(a^128+b^128)(a^256+b^256)(a^512+b^512)(a^1024+b^1024)

    = (a^64-b^64)(a^64+b^64)(a^128+b^128)(a^256+b^256)(a^512+b^512)(a^1024+b^1024)

    = (a^128-b^128)(a^128+b^128)(a^256+b^256)(a^512+b^512)(a^1024+b^1024)

    = (a^256-b^256)(a^256+b^256)(a^512+b^512)(a^1024+b^1024)

    = (a^512-b^512)(a^512+b^512)(a^1024+b^1024)

    = (a^1024-b^1024)(a^1024+b^1024)

    = (a^1024)^2 - (b^1024)^2

    = a^2048 - b^2048

    =============================

    Simpler way :

    (a-b)(a+b)(a^2+b^2)(a^4+b^4)(a^8+b^8)(a^16+b^16).....(a^1024+b^1024)

    Note that (a-b)(a+b) = a^2 - b^2 and we know (a^2-b^2)(a^2+b^2) = (a^2)^2 - (b^2)^2 = a^4 - b^4 , also (a^4-b^4)(a^4+b^4) = (a^4)^2 - (b^4)^2 = a^8 - b^8 ...

    Do first two steps :

    (a-b)(a+b)(a^2+b^2)(a^4+b^4)(a^8+b^8)(a^16+b^16).....(a^1024+b^1024)

    = (a^2-b^2)(a^2+b^2)(a^4+b^4)(a^8+b^8)(a^16+b^16).....(a^1024+b^1024)

    = (a^4-b^4)(a^4+b^4)(a^8+b^8)(a^16+b^16).....(a^1024+b^1024)

    see the yellow , so we know we can simplify into :

    (a^1024)^2 - (b^1024)^2 = a^(2×1024) - b^(2×1024) = a^2048 - b^2048

    =============================

    To conclude , we get the answer :

    (a-b)(a+b)(a^2+b^2)(a^4+b^4)(a^8+b^8)(a^16+b^16).....(a^1024+b^1024) =

    a^2048 - b^0248

    2011-07-19 20:40:22 補充:

    To conclude , we get the answer :

    (a-b)(a+b)(a^2+b^2)(a^4+b^4)(a^8+b^8)(a^16+b^16).....(a^1024+b^1024) =

    a^2048 - b^2048

    資料來源: Hope I Can Help You ^_^
  • Lll
    Lv 5
    10 年前

    timmy的答案應該不對

  • 10 年前

    (a-b)(a+b)=a^2-b^2

    (a^2+b^2)(a^2-b^2)=a^4-b^4

    .

    .

    .

    so (a-b)(a+b)(a^2+b^2)(a^4+b^4)(a^8+b^8)(a^16+b^16).....(a^1024+b^1024)

    =(a^1024-b^1024)(a^1024+b^1024)

    =a^2048-b^2048

    資料來源: 普通數學理論
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