7.Binomial Theorem
hard

यदि ${(1 + x)^n}$ के प्रसार में चार क्रमिक पदों के गुणांक ${a_1},{a_2},{a_3},{a_4}$ हैं, तब $\frac{{{a_1}}}{{{a_1} + {a_2}}} + \frac{{{a_3}}}{{{a_3} + {a_4}}}$=

A

$\frac{{{a_2}}}{{{a_2} + {a_3}}}$

B

$\frac{1}{2}\frac{{{a_2}}}{{({a_2} + {a_3})}}$

C

$\frac{{2{a_2}}}{{{a_2} + {a_3}}}$

D

$\frac{{2{a_3}}}{{{a_2} + {a_3}}}$

(IIT-1975)

Solution

यदि ${(1 + x)^n}$ के प्रसार में $(r + 1)$वे,$(r + 2)$वे,$(r + 3)$वे तथा $(r + 4)$वे पदों के गुणांक क्रमश: ${a_1},{a_2},{a_3},{a_4}$ है

तब ${a_1} = {\,^n}{C_r},{a_2} = {\,^n}{C_{r + 1}},{a_3} = {\,^n}{C_{r + 2,}}{a_4} = {\,^n}{C_{r + 3}}$

अब $\frac{{{a_1}}}{{{a_1} + {a_2}}} + \frac{{{a_3}}}{{{a_3} + {a_4}}} = \frac{{^n{C_r}}}{{^n{C_r} + {\,^n}{C_{r + 1}}}}$$ + \frac{{^n{C_{r + 2}}}}{{^n{C_{r + 2}} + {\,^n}{C_{r + 3}}}}$

$ = \frac{{^n{C_r}}}{{^{n + 1}{C_{r + 1}}}} + \frac{{^n{C_{r + 2}}}}{{^{n + 1}{C_{r + 3}}}}$$ = \frac{{^n{C_r}}}{{\frac{{n + 1}}{{r + 1}}{\,^n}{C_r}}} + \frac{{^n{C_{r + 2}}}}{{\frac{{n + 1}}{{r + 3}}{\,^n}{C_{r + 2}}}}$

$ = \frac{{r + 1}}{{n + 1}} + \frac{{r + 3}}{{n + 1}} = \frac{{2(r + 2)}}{{n + 1}}$

$ = 2\frac{{^n{C_{r + 1}}}}{{^{n + 1}{C_{r + 2}}}} = 2\frac{{^n{C_{r + 1}}}}{{^n{C_{r – 1}} + {\,^n}{C_{r + 2}}}} = \frac{{2{a_2}}}{{{a_2} + {a_3}}}$

Standard 11
Mathematics

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