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RBSE Solutions for Class 11 Maths Chapter 7 द्विपद प्रमेय Ex 7.3

June 12, 2019 by Prasanna Leave a Comment

Rajasthan Board RBSE Class 11 Maths Chapter 7 द्विपद प्रमेय Ex 7.3

प्रश्न 1.
यदि (1 + x)n के प्रसार के गुणांक क्रमशः C0, C1, C2, ………. Cn हो, तो मान ज्ञात कीजिए-
(i) 8C1 + 8C2 + 8C3 + …… 8C8
(ii) 8C1 + 8C3 + 8C5 + 8C7 + ……..
हल-
(i) चूँकि हम जानते हैं कि
nC0 + nC1 + nC2 + ….. + nCn = 2n
1 + nC1 + nC2 + nC3 +….. + nCn = 2n
∴ nC1 + nC2 + nC3 + ….. + nCn = 2n – 1
n = 8 रखने पर
8C1 + 8C2 + 8C3 + ….. + 8C8 = 28 – 1
= 256 – 1 = 255

(ii) द्विपद गुणांकों के गुणधर्म से हम जानते हैं ।
2(nC1 + nC3 + nC5 + ……) = 2n
या nC1 + nC3 + nC5 + …. \(\frac { { 2 }^{ n } }{ 2 } \) = 2n-1
n = 8 रखने पर
8C1 + 8C3 + 8C5 + …… = 28-1 = 27
= 128
यदि (1 + x)n के प्रसार के गुणांक क्रमशः C0, C1, C2 ,…… Cn हो, तब सिद्ध कीजिए

प्रश्न 2.
C0 + 3.C1 + 5.C2 +……. + (2n + 1). Cn = (n + 1)2n
हल-
L.H.S. C0 + 3.C1 + 5.C2 + ……. + (2n + 1) . Cn
= (C0 + C1 + C2 + C3 + ….. + Cn) + (2.C1 + 4.C2 + 6.C3 + …… + 2nCn)
= 2n + 2(C1 + 2C2 + 3C3 …… + n.Cn).
RBSE Solutions for Class 11 Maths Chapter 7 द्विपद प्रमेय Ex 7.3
= 2n + 2n[1 + n-1C1 + n-1C2 + n-1C3 +….. n-1Cn-1].
= 2n + 2n.2n-1
= 2n + n.2n
= (n + 1)2n
= R.H.S.
∴ L.H.S. = R.H.S.

प्रश्न 3.
C0C2 + C1C3 + C2C4 +…… + Cn-2Cn =
RBSE Solutions for Class 11 Maths Chapter 7 द्विपद प्रमेय Ex 7.3
हल :
(1 + x)n = (C0 + C1x + C2x2 + C3x3 + ….. + Cnxn) ….(1)
(x + 1)n = (C0xn + C1xn-1 + C2xn-2 + …… + Cn) …(2)
समी. (1) तथा (2) का गुणा करने पर
(1 + x)n × (x + 1)n
= (C0 + C1x + C2x2 + C3x3 + ….. + Cnxn) ×
(C0xn + C1xn-1 + C2xn-2 + …… + Cn)
दोनों पक्षों में xn-r के गुणांकों की तुलना करने पर
2nCn-r = C0Cr + C1rr+1 + …… + Cn-r Cr
r = 2 रखने पर
2nCn-2 = C0C2 + C1C3 + C2C4 …… Cn-2Cn
RBSE Solutions for Class 11 Maths Chapter 7 द्विपद प्रमेय Ex 7.3

प्रश्न 4.
C0 + 2C1 + 4C2 + 6C3 + ….. + 2nCn = 1 + n2n
हल-
L.H.S.
C0 + 2C1 + 4C2 + 6C3 + ….. + 2nCn
= 1 + 2C1 + 4C2 + 6C3 + …… + 2nCn
∵ C0 = 1
= 1 + 2(C1 + 2C2 + 3C3 +…….. + nCn)
= 1 + 2(nC1 + 2.nC2 + 3.nC3+ ……. + n.nCn)
RBSE Solutions for Class 11 Maths Chapter 7 द्विपद प्रमेय Ex 7.3

प्रश्न 5.
RBSE Solutions for Class 11 Maths Chapter 7 द्विपद प्रमेय Ex 7.3
हल-
सभी के मान निकालने पर
RBSE Solutions for Class 11 Maths Chapter 7 द्विपद प्रमेय Ex 7.3

प्रश्न 6.
यदि (1 + x – 2x2)6 का पूर्ण प्रसार 1 + a1x + a2x2 + a3x3 + …….. + a12x12 द्वारा निरूपित हो, तब सिद्ध कीजिए, a2 + a4 + a6 + ……… a12 = 31
हल-
x = -1 रखने पर
(1 – 1 – 2(-1)2)6 = 1 – a1 + a2 – a3 + ….. + a12
अर्थात् (-2)6 = 1 – a1 + a2 – a3 + a4 – a5 + …… + a12
64 = 1 – a1 + a2 – a3 + a4 – a5 + …… + a12 ….(1)
अब x = 1 रखने पर
(1 + 1 – 2.12)6 = 1 + a1 + a2 + a3 + ……. + a12
(2 – 2)6 = 1 + a1 + a2 + a3 + …….. a12
0 = 1 + a1 + a2 + a3 + ……. + a12 ….(2)
समीकरण (1) तथा (2) को जोड़ने पर
64 = 2 + 2a2 + 2a4 + …… + 2a12
64 = 2(1 + a2 + a4 + …… + a12)
या \(\frac { 64 }{ 2 }\) = (1 + a2 + a4 + …… + a12)
32 = 1 + a2 + a4 + ….. + a12
या a2 + a4 + a6 + ……. + a12 = 32 – 1 = 31
अतः a2 + a4 + a6 + ……. + a12 = 31 इतिसिद्धम्

RBSE Solutions for Class 11 Maths

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