Let C _{ r } denote binomial coefficient of x ^{ r } in expansion of (1+x)^{10} . If α, β ∈ R .

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7.Binomial Theorem

normal

Let $C _{ r }$ denote the binomial coefficient of $x ^{ r }$ in the expansion of $(1+x)^{10}$. If $\alpha, \beta \in R$. $C _{1}+3 \cdot 2 C _{2}+5 \cdot 3 C _{3}+\ldots$ upto $10$ terms $=\frac{\alpha \times 2^{11}}{2^{\beta}-1}\left( C _{0}+\frac{ C _{1}}{2}+\frac{ C _{2}}{3}+\ldots . .\right.$ upto 10 terms $)$ then the value of $\alpha+\beta$ is equal to

BONUS

BONUS PLUSE

PLUSE

PLUSE BONUE

(JEE MAIN-2022)

Solution

$(1+x)^{10}=C_{0}+C_{1} x+C_{2} x^{2}+\ldots \ldots+C_{10} x^{10}$

Differentiating

$10(1+x)^{9}=C_{1}+2 C_{2} x+3 C_{3} x^{2}+\ldots+10 C_{10} x^{9}$

replace $x \rightarrow X ^{2}$

$10\left(1+x^{2}\right)^{9}=C_{1}+2 C_{2} x^{2}+3 C_{3} x^{4}+\ldots+10 C_{10} x^{18}$

$10 \cdot x\left(1+x^{2}\right)^{9}=C_{1} x+2 C_{2} x^{3}+3 C_{3} x^{5}+\ldots .+10 C_{10} x^{19}$

Differentiating

$10\left(\left(1+x^{2}\right)^{9} \cdot 1+x \cdot 9\left(1+x^{2}\right)^{8} 2 x\right)$

$=C_{1} x+2 C_{2} \cdot 3 x^{3}+3 \cdot 5 \cdot C_{3} x^{4}+\ldots .+10 \cdot 19 C_{10} x^{18}$

putting $x=1$

$10\left(2^{9}+18 \cdot 2^{8}\right)$

$= C _{1}+3 \cdot 2 \cdot C _{2}+5 \cdot 3 \cdot C _{3}+\ldots+19 \cdot 10 \cdot C _{10} $

$C _{1}+3 \cdot 2 \cdot C _{2}+\ldots \ldots+19 \cdot 10 \cdot C _{10}$

$=10 \cdot 2^{9} \cdot 10=100 \cdot 2^{9}$

$C _{0}+\frac{ C _{1}}{2}+\frac{ C _{2}}{3}+\ldots . .+\frac{ C _{9}}{11}+\frac{ C _{10}}{11}=\frac{2^{11}-1}{11}$

$10^{\text {th }} \text { term } 11^{\text {th }} \text { term }$

$C _{0}+\frac{ C _{1}}{2}+\frac{ C _{2}}{3}+\ldots .+\frac{ C _{9}}{11}=\frac{2^{11}-2}{11}$

Now, $100 \cdot 2^{9}=\frac{\alpha \cdot 2^{11}}{2^{\beta}-1}\left(\frac{2^{11}-2}{11}\right)$

Eqn. of form $y = k \left(2^{ x }-1\right)$.

It has infinite solutions even if we take $x, y \in N$.

Standard 11

Mathematics

If $1+\left(2+{ }^{49} C _{1}+{ }^{49} C _{2}+\ldots .+{ }^{49} C _{49}\right)\left({ }^{50} C _{2}+{ }^{50} C _{4}+\right.$ $\ldots . .+{ }^{50} C _{ so }$ ) is equal to $2^{ n } . m$, where $m$ is odd, then $n$ $+m$ is equal to.

hard

(JEE MAIN-2022)

Let n and k be positive integers such that $n \ge \frac{{k(k + 1)}}{2}$. The number of solutions $({x_1},{x_2},….{x_k})$, ${x_1} \ge 1,{x_2} \ge 2,….{x_k} \ge k,$ all integers, satisfying ${x_1} + {x_2} + …. + {x_k} = n$, is

normal

(IIT-1996)

If $A$ denotes the sum of all the coefficients in the expansion of $\left(1-3 x+10 x^2\right)^n$ and $B$ denotes the sum of all the coefficients in the expansion of $\left(1+x^2\right)^n$, then :

medium

(JEE MAIN-2024)

If ${(1 + x)^n} = {C_0} + {C_1}x + {C_2}{x^2} + …. + {C_n}{x^n}$, then the value of ${C_0} + 2{C_1} + 3{C_2} + …. + (n + 1){C_n}$ will be

medium

(IIT-1971)

If $\sum_{r=1}^{10} r !\left( r ^{3}+6 r ^{2}+2 r +5\right)=\alpha(11 !),$ then the value of $\alpha$ is equal to …… .