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A chruthú leis an ionduchtú go bhfuil (1 + x) n ≥ 1 + nx for x > -1, n  N

A chruthú leis an ionduchtú go bhfuil (1 + x) n ≥ 1 + nx for x > -1, n  N. NB x>-1. Dá réir sin (1+x)>0. Cruthaigh go bhfuil sé fíor i gcás n = 1. n = 1 (1 + x) 1 = 1 + x. Is fíor é i gcás n = 1. Glac leis gur fíor é i gcás n = k Dá bhrí sin (1 + x) k ≥ 1 + kx.

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A chruthú leis an ionduchtú go bhfuil (1 + x) n ≥ 1 + nx for x > -1, n  N

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  1. A chruthú leis an ionduchtú go bhfuil (1 + x)n ≥ 1 + nx for x > -1, n  N NB x>-1. Dá réir sin (1+x)>0. Cóipcheart Foireann Fhorbartha Thionscadal Mata 2012

  2. Cruthaigh go bhfuil sé fíor i gcás n = 1 n = 1 (1 + x)1 = 1 + x Is fíor é i gcás n = 1 Glac leis gur fíor é i gcás n = k Dá bhrí sin (1 + x)k ≥ 1 + kx Cóipcheart Foireann Fhorbartha Thionscadal Mata 2012

  3. Glac leis gur fíor é i gcás n = k + 1 Iolraigh an Dá Thaobh faoi 1 + x (1 + x)(1 + x)k ≥ (1 + x)(1 + kx) (1 + x)k+1 ≥ 1 + kx + x + kx2 Mura bhuil (1 + x)k+1 ≥ 1 + kx + x + kx2 (k > 0 → kx2 ≥0 do gac x) Dá bhrí sin (1 + x)k+1 ≥ 1 + kx + x (1 + x)k+1 ≥ 1 + (k+1)x Má tá sé fíor i gcás n = k, tugann sé sin le fios gur fíor é i gcás n = k + 1 Dá réir sin leis an Ionduchtú (1 + x)n ≥ 1 + nx x > -1, n  N. Cóipcheart Foireann Fhorbartha Thionscadal Mata 2012

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