Question

English

Gujarati

Hindi

7.Binomial Theorem

hard

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 $\frac{{{a_1}}}{{{a_1} + {a_2}}} + \frac{{{a_3}}}{{{a_3} + {a_4}}}$ =

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

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

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

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

(IIT-1975)

Solution

(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}}}$

Std 11

Mathematics

Let $(1+2 x)^{20}=a_0+a_1 x+a_2 x^2+\ldots+a_{20} x^{20}$.Then $3 a_0+2 a_1+3 a_2+2 a_3+3 a_4+2 a_5+\ldots+2 a_{19}+3 a_{20}$ equals

normal

The value of$^n{C_1}\sum\limits_{r = 0}^1 {^1{C_r}} { + ^n}{C_2}\left( {\sum\limits_{r = 0}^2 {^2{C_r}} } \right){ + ^n}{C_3}\left( {\sum\limits_{r = 0}^3 {^3{C_r}} } \right) + ……{ + ^n}{C_n}\left( {\sum\limits_{r = 0}^n {^n{C_r}} } \right)$ is equal to

normal

The sum to $(n + 1)$ terms of the series $\frac{{{C_0}}}{2} – \frac{{{C_1}}}{3} + \frac{{{C_2}}}{4} – \frac{{{C_3}}}{5} + …$ is

normal

If ${\sum\limits_{i = 1}^{20} {\left( {\frac{{{}^{20}{C_{i – 1}}}}{{{}^{20}{C_i} + {}^{20}{C_{i – 1}}}}} \right)} ^3}\, = \frac{k}{{21}}$, then $k$ equals

hard

The coefficient of $t^{50}$ in $(1 + t^2)^{25} (1 + t^{25}) (1 + t^{40}) (1 + t^{45}) (1 + t^{47})$ is

normal

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 $\frac{{{a_1}}}{{{a_1} + {a_2}}} + \frac{{{a_3}}}{{{a_3} + {a_4}}}$ =

Solution

Similar Questions

Let $(1+2 x)^{20}=a_0+a_1 x+a_2 x^2+\ldots+a_{20} x^{20}$.Then $3 a_0+2 a_1+3 a_2+2 a_3+3 a_4+2 a_5+\ldots+2 a_{19}+3 a_{20}$ equals

The value of$^n{C_1}\sum\limits_{r = 0}^1 {^1{C_r}} { + ^n}{C_2}\left( {\sum\limits_{r = 0}^2 {^2{C_r}} } \right){ + ^n}{C_3}\left( {\sum\limits_{r = 0}^3 {^3{C_r}} } \right) + ……{ + ^n}{C_n}\left( {\sum\limits_{r = 0}^n {^n{C_r}} } \right)$ is equal to

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The coefficient of $t^{50}$ in $(1 + t^2)^{25} (1 + t^{25}) (1 + t^{40}) (1 + t^{45}) (1 + t^{47})$ is