If a 1 , a 2 , a 3 , a 4 are the coefficients of any four consecutive terms in the expansion of ( 1 + x ) n , then a 1 a 1 + a 2 + a 3 a 3 + a 4 = Options: A . &nbsp a 2 a 2 + a 3 B . &nbsp 1 2 a 2 a 2 + a 3 C . &nbsp 2 a 2 a 2 + a 3 D . &nbsp 2 a 3 a 2 + a 3

Answer: Option C : 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, a 1 a 1 + a 2 + a 3 a 3 + a 4 = n C r n C r + n C r + 1 + n C r + 2 n C r + 2 + n C r + 3 = n C r n + 1 C r + 1 + n C r + 2 n + 1 C r + 3 = n C r n + 1 r + 1 n C r + n C r n + 1 r + 3 n C r + 2 = r + 1 n + 1 + r + 3 n + 3 = 2 ( r + 2 ) n + 1 Also, solving the R.H.S, we get 2 a 2 a 2 + a 3 = 2 n C r + 1 n C r + 1 + n C r + 2 = 2 n C r + 1 n + 1 C r + 2 = 2 ( r + 2 ) n + 1

If a1,a2,a3,a4 are the coefficients of any four consecutive terms in the

Discussion Forum : Binomial Theorem

Question -

If $a_{1}, a_{2}, a_{3}, a_{4}$ are the coefficients of any four consecutive terms in the expansion of $(1 + x)^{n}$ , then $\frac{a_{1}}{a_{1} + a_{2}} + \frac{a_{3}}{a_{3} + a_{4}} =$

Options:

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}}

Answer: Option C
:
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,
$\frac{a_{1}}{a_{1} + a_{2}} + \frac{a_{3}}{a_{3} + a_{4}} = \frac{{^{n} C}_{r}}{{^{n} C}_{r} + {^{n} C}_{r + 1}} + \frac{{^{n} C}_{r + 2}}{{^{n} C}_{r + 2} + {^{n} C}_{r + 3}} = \frac{{^{n} C}_{r}}{{^{n + 1} C}_{r + 1}} + \frac{{^{n} C}_{r + 2}}{{^{n + 1} C}_{r + 3}} = \frac{{^{n} C}_{r}}{\frac{n + 1}{r + 1} {^{n} C}_{r}} + \frac{{^{n} C}_{r}}{\frac{n + 1}{r + 3} {^{n} C}_{r + 2}} = \frac{r + 1}{n + 1} + \frac{r + 3}{n + 3} = \frac{2 (r + 2)}{n + 1}$
Also, solving the R.H.S, we get
$\frac{2 a_{2}}{a_{2} + a_{3}} = 2 \frac{{^{n} C}_{r + 1}}{{^{n} C}_{r + 1} + {^{n} C}_{r + 2}} = 2 \frac{{^{n} C}_{r + 1}}{{^{n + 1} C}_{r + 2}} = \frac{2 (r + 2)}{n + 1}$

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