જો {a_1},{a_2},{a_3},{a_4} એ {(1 + x)^n} ની વ | vedclass

Name: PDF Generator App | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(c) Let ${a_1},{a_2},{a_3},{a_4}$ be respectively the coefficients of ${(r + 1)^{th}},{(r + 2)^{th}}$, ${(r + 3)^{th}}$ and ${(r + 4)^{th}}$terms in t

Vedclass · Accepted Answer

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

Vedclass · Answer

(c) Let ${a_1},{a_2},{a_3},{a_4}$ be respectively the coefficients of ${(r + 1)^{th}},{(r + 2)^{th}}$, ${(r + 3)^{th}}$ and ${(r + 4)^{th}}$terms in the expansion of ${(1 + x)^n}$.

Then ${a_1} = {\,^n}{C_r},{a_2} = {\,^n}{C_{r + 1}},{a_3} = {\,^n}{C_{r + 2,}}{a_4} = {\,^n}{C_{r + 3}}$

Now $\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}}}}$
                                               

                                                   $({^n}{C_r} = \frac{n}{r}{\,^{n - 1}}{C_{r - 1}})$

= $\frac{{r + 1}}{{n + 1}} + \frac{{r + 3}}{{n + 1}} = \frac{{2(r + 2)}}{{n + 1}}$

$ = 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}}}$

જો ${a_1},{a_2},{a_3},{a_4}$ એ ${(1 + x)^n}$ ની વિસ્તરણના ચાર ક્રમિક પદ હોય , તો $\frac{{{a_1}}}{{{a_1} + {a_2}}} + \frac{{{a_3}}}{{{a_3} + {a_4}}}$ =

Similar Questions

જો $(1 - 2x + 5x^2 - 10x^3) (1 + x)^n = 1 + a_1x + a_2x^2 + ....$ આપેલ હોય અને $a_1^2\,= 2a_2$ હોય તો $n$ ની કિમત મેળવો

$\sum\limits_{n = 1}^\infty {\frac{{^n{C_0} + ...{ + ^n}{C_n}}}{{^n{P_n}}}} $ = . . .

$(1+x)^{500}+x(1+x)^{499}+x^2(1+x)^{498}+\ldots . .+x^{500}$ માં $x ^{301}$નો સહગુણક $........$ છે.

જો ${(x - 2y + 3z)^n}$ ના વિસ્તરણમાં પદની સંખ્યા $45$ હોય , તો $n= $. . .