(1 + x)2n के विस्तार में पहले और अंतिम पदों के गुणांक का योग क्या है, जहां n एक प्राकृतिक संख्या है?

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  1. 1
  2. 2
  3. n
  4. 2n

Answer (Detailed Solution Below)

Option 2 : 2
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अवधारणा:

\(\rm ^n C_r = {n!\over(r!(n - r)!)}\)

(1 + x)n = nC0 × 1(n-0) × x 0nC1 × 1(n-1) × x 1 + nC2  × 1(n-2) × x2 + …. + nCn  × 1(n-n) × xn

 

गणना:

दिया गया विस्तार (1 + x)2n है

2nC×1(2n-0) × x0 +  2nC1 ×1(2n-1) × x1 + ... +  2nC2n ×1(2n-2n) × x2n

पहला पद = 2nC×1 × 1 = 1

अंतिम पद =  2nC2n ×1 × x2n = 1 × x2n = x2n

योग = 1 + x2n

1 का गुणांक = 1, x2n का गुणांक = 1

∴  तो, गुणांकों का योग = 1 + 1 = 2

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