के विस्तार में अंतिम से 4वां पद ज्ञात कीजिए।

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Navik GD Mathematics 22 March 2021 (All Shifts) Questions
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  1. 108x3y4
  2. 105x3y5
  3. 105x3y4
  4. 105x2y5

Answer (Detailed Solution Below)

Option 3 : 105x3y4
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अवधारणा:

द्विपद विस्तार:

  • (a + b)n = Can b+ Can-1 b1 + Can-2 b2 + … + Can-r br + … + Cn-1 a1 bn-1 + Cn a0 bn, जहाँ C0, C1, …, Cn, Cr = nCr =  के रूप में परिभाषित द्विपद गुणांक हैं
  • विस्तार में पदों की कुल संख्या n + 1 है।
  • विस्तार में (r + 1)वां पद Tr+1 = Cr an-r br है।

 

गणना:

दिए गए व्यंजक के लिए , n = 7

उपरोक्त व्यंजक के विस्तार में 7 + 1 = 8 पद होंगे।

पदों के क्रम को उलटने पर विस्तार करके (अर्थात a = -3y और b = ), हमारे पास है:

T5 = T4+1 = 7C4  (-3y)4  = 35  (34y4 = 105x3y4.

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