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What is the last number of terms of a G.P. 2+4+8 ... that will give a sum greater than 1,500,000?

Mathematics

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Answer

20

Step-by-step solution

6 steps
  1. Tambua neno la kwanza (aa) na uwiano wa kawaida (rr) wa GP.

    Kutoka kwa mfuatano 2,4,8,…2, 4, 8, \dots:
    Neno la kwanza, a=2a = 2.
    Uwiano wa kawaida, r=42=2r = \frac{4}{2} = 2.

  2. Andika fomula ya jumla ya maneno nn ya GP.

    Kwa kuwa r>1r > 1, fomula ya jumla ya maneno nn (SnS_n) ni:

    Sn=a(rn−1)r−1S_n = \frac{a(r^n - 1)}{r - 1}
  3. Badilisha thamani za aa na rr kwenye fomula.

    Sn=2(2n−1)2−1S_n = \frac{2(2^n - 1)}{2 - 1} Sn=2(2n−1)1S_n = \frac{2(2^n - 1)}{1} Sn=2(2n−1)S_n = 2(2^n - 1)
  4. Weka usawa kulingana na sharti kwamba jumla ni kubwa kuliko 1,500,000.

    2(2n−1)>1,500,0002(2^n - 1) > 1,500,000
  5. Tatua usawa kwa nn.

    Gawanya pande zote mbili kwa 2:

    2(2n−1)2>1,500,0002\frac{2(2^n - 1)}{2} > \frac{1,500,000}{2} 2n−1>750,0002^n - 1 > 750,000

    Ongeza 1 kwa pande zote mbili:

    2n>750,000+12^n > 750,000 + 1 2n>750,0012^n > 750,001

    Ili kutatua kwa nn, chukua logariti ya pande zote mbili (kwa mfano, logariti ya msingi 10 au asili):

    log⁡(2n)>log⁡(750,001)\log(2^n) > \log(750,001) nlog⁡(2)>log⁡(750,001)n \log(2) > \log(750,001) n>log⁡(750,001)log⁡(2)n > \frac{\log(750,001)}{\log(2)}

    Tumia kikokotoo kupata thamani za logariti:

    log⁡(750,001)≈5.87506\log(750,001) \approx 5.87506 log⁡(2)≈0.30103\log(2) \approx 0.30103 n>5.875060.30103n > \frac{5.87506}{0.30103} n>19.516n > 19.516
  6. Tafuta nambari kamili ndogo zaidi ya maneno nn inayokidhi usawa.

    Kwa kuwa nn lazima iwe nambari kamili (idadi ya maneno), na nn lazima iwe kubwa kuliko 19.516, nambari kamili ndogo zaidi inayokidhi sharti hili ni 20.

    Nambari ndogo zaidi ya maneno ambayo itatoa jumla kubwa kuliko 1,500,000 ni 20.

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