What is the value of ${ }^{23} C_0+{ }^{23} C_2+{ }^{23} C_4+\cdots+{ }^{23} C_{22}$ ?...

$$ (1+x)^n={ }^n C_0+{ }^n C_1 x+\cdots+{ }^n C_n x^n $$Let $\mathrm{x}=1$$$ \begin{gathered} \Rightarrow C_0+C_1 \cdot 1+C_2 \cdot 1^2+\cdots+C_n \cdot 1^n=(1+1)^n=2^n \\ \Rightarrow C_0+C_1+C_2 \cdot 1+\cdots+C_n=(1+1)^n=2^n \end{gathered} $$So, ${ }^{23} C_0+{ }^{23} C_2+{ }^{23} C_4+\cdots+{ }^{23} C_{22}=2^{22}$Given,$$ \begin{gathered} f(x+y)=f(x)+f(y) \\ f(1+1)=f(1)+f(1) \rightarrow f(2)=20 \\ f(1+1)=f(1)+f(1+1)=3 f(1) \rightarrow f(3)=30 \end{gathered} $$Clearly, $f(x)=10 x$ which is one-and onto

Let $n=\sum\limits_{r=0}^{10}(-1)^r{ }^{10} C_r\left(\frac{2}{3}\right)^{2 r} 3^{20}$. Which one of ...

Let's restate the problem carefully:$ n = \sum_{r=0}^{10} (-1)^r {10 \choose r} \left( \frac{2}{3} \right)^{2r} 3^{20} $We are asked to determine which of the given statements about divisibility istrue.Step 1: Simplify the power expression$ \left( \frac{2}{3} \right)^{2r} = \frac{2^{2r}}{3^{2r}} = \frac{4^r}{3^{2r}} $So,$ n = 3^{20} \sum_{r=0}^{10} (-1)^r {10 \choose r} \frac{4^r}{3^{2r}} $$ n = 3^{20} \sum_{r=0}^{10} {10 \choose r} (-1)^r \left( \frac{4}{9} \right)^r $Step 2: Recognize as a binomial expansion$ (1 - x)^{10} = \sum_{r=0}^{10} (-1)^r {10 \choose r} x^r $Compare: here $ x = \frac{4}{9} $$ \implies \sum_{r=0}^{10} (-1)^r {10 \choose r} \left( \frac{4}{9} \right)^r = (1 - \frac{4}{9})^{10} = \left( \frac{5}{9} \right)^{10} $Step 3: Substitute back$ n = 3^{20} \left( \frac{5}{9} \right)^{10} = 3^{20} \cdot \frac{5^{10}}{9^{10}} $But $ 9^{10} = (3^2)^{10} = 3^{20} $.So:$ n = 3^{20} \cdot \frac{5^{10}}{3^{20}} = 5^{10} $Step 4: Now we know $ n = 5^{10} $Let's check divisibility:$5^{10}$ is divisible by5Not divisible by 6 (needs factor 2,3)Not divisible by 8 (no factor 2)Not divisible by 9 (no factor 3)✅Therefore, $n$ is divisible by 5 only.✅Final Answer: Option A — $n$ is divisible by 5.

Binomial Theorem - IAT-IISER Maths PYQs

(1%2Bx)%5En%3D%7B%20%7D%5En%20C_0%2B%7B%20%7D%5En%20C_1%20x%2B%5Ccdots%2B%7B%20%7D%5En%20C_n%20x%5En

Let

\mathrm{x}=1$$

%5Cbegin%7Bgathered%7D%20%5CRightarrow%20C_0%2BC_1%20%5Ccdot%201%2BC_2%20%5Ccdot%201%5E2%2B%5Ccdots%2BC_n%20%5Ccdot%201%5En%3D(1%2B1)%5En%3D2%5En%20%5C%5C%20%5CRightarrow%20C_0%2BC_1%2BC_2%20%5Ccdot%201%2B%5Ccdots%2BC_n%3D(1%2B1)%5En%3D2%5En%20%5Cend%7Bgathered%7D

So%2C

{ }^{23} C_0+{ }^{23} C_2+{ }^{23} C_4+\cdots+{ }^{23} C_{22}=2^{22}

Given%2C

%5Cbegin%7Bgathered%7D%20f(x%2By)%3Df(x)%2Bf(y)%20%5C%5C%20f(1%2B1)%3Df(1)%2Bf(1)%20%5Crightarrow%20f(2)%3D20%20%5C%5C%20f(1%2B1)%3Df(1)%2Bf(1%2B1)%3D3%20f(1)%20%5Crightarrow%20f(3)%3D30%20%5Cend%7Bgathered%7D

Clearly%2C

f(x)=10 x$ which is one-and onto

Let's restate the problem carefully:

n%20%3D%20%5Csum_%7Br%3D0%7D%5E%7B10%7D%20(-1)%5Er%20%7B10%20%5Cchoose%20r%7D%20%5Cleft(%20%5Cfrac%7B2%7D%7B3%7D%20%5Cright)%5E%7B2r%7D%203%5E%7B20%7D

We are asked to determine which of the given statements about divisibility istrue.Step 1: Simplify the power expression

%5Cleft(%20%5Cfrac%7B2%7D%7B3%7D%20%5Cright)%5E%7B2r%7D%20%3D%20%5Cfrac%7B2%5E%7B2r%7D%7D%7B3%5E%7B2r%7D%7D%20%3D%20%5Cfrac%7B4%5Er%7D%7B3%5E%7B2r%7D%7D

So,

n%20%3D%203%5E%7B20%7D%20%5Csum_%7Br%3D0%7D%5E%7B10%7D%20(-1)%5Er%20%7B10%20%5Cchoose%20r%7D%20%5Cfrac%7B4%5Er%7D%7B3%5E%7B2r%7D%7D

n%20%3D%203%5E%7B20%7D%20%5Csum_%7Br%3D0%7D%5E%7B10%7D%20%7B10%20%5Cchoose%20r%7D%20(-1)%5Er%20%5Cleft(%20%5Cfrac%7B4%7D%7B9%7D%20%5Cright)%5Er

Step 2: Recognize as a binomial expansion

(1%20-%20x)%5E%7B10%7D%20%3D%20%5Csum_%7Br%3D0%7D%5E%7B10%7D%20(-1)%5Er%20%7B10%20%5Cchoose%20r%7D%20x%5Er

Compare: here

x%20%3D%20%5Cfrac%7B4%7D%7B9%7D

%5Cimplies%20%5Csum_%7Br%3D0%7D%5E%7B10%7D%20(-1)%5Er%20%7B10%20%5Cchoose%20r%7D%20%5Cleft(%20%5Cfrac%7B4%7D%7B9%7D%20%5Cright)%5Er%20%3D%20(1%20-%20%5Cfrac%7B4%7D%7B9%7D)%5E%7B10%7D%20%3D%20%5Cleft(%20%5Cfrac%7B5%7D%7B9%7D%20%5Cright)%5E%7B10%7D

Step 3: Substitute back

n%20%3D%203%5E%7B20%7D%20%5Cleft(%20%5Cfrac%7B5%7D%7B9%7D%20%5Cright)%5E%7B10%7D%20%3D%203%5E%7B20%7D%20%5Ccdot%20%5Cfrac%7B5%5E%7B10%7D%7D%7B9%5E%7B10%7D%7D

But

9%5E%7B10%7D%20%3D%20(3%5E2)%5E%7B10%7D%20%3D%203%5E%7B20%7D

.So:

n%20%3D%203%5E%7B20%7D%20%5Ccdot%20%5Cfrac%7B5%5E%7B10%7D%7D%7B3%5E%7B20%7D%7D%20%3D%205%5E%7B10%7D

Step 4: Now we know

n%20%3D%205%5E%7B10%7D

Let's check divisibility:

5%5E%7B10%7D

is divisible by5Not divisible by 6 (needs factor 2,3)Not divisible by 8 (no factor 2)Not divisible by 9 (no factor 3)✅Therefore,

n

is divisible by 5 only.✅Final Answer: Option A —

n

is divisible by 5.

Binomial Theorem

High-Yield Trend

Questions
2 MCQs

Official Solution

Official Solution

Binomial Theorem

High-Yield Trend

Questions 2 MCQs

Official Solution

Official Solution

Questions
2 MCQs