Question
Download Solution PDF(1 + x)2n के विस्तार में पहले और अंतिम पदों के गुणांक का योग क्या है, जहां n एक प्राकृतिक संख्या है?
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFअवधारणा:
\(\rm ^n C_r = {n!\over(r!(n - r)!)}\)
(1 + x)n = nC0 × 1(n-0) × x 0+ nC1 × 1(n-1) × x 1 + nC2 × 1(n-2) × x2 + …. + nCn × 1(n-n) × xn
गणना:
दिया गया विस्तार (1 + x)2n है
= 2nC0 ×1(2n-0) × x0 + 2nC1 ×1(2n-1) × x1 + ... + 2nC2n ×1(2n-2n) × x2n
पहला पद = 2nC0 ×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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