Math, asked by mgssksinha, 1 year ago

evaluate 22^8/(1+3+5+7+9+...............239+241)^2​

Answers

Answered by shadowsabers03
2

             

\Rightarrow\ \boxed{\frac{22^8}{(1+3+5+......+241)^2}} \\ \\ \\ \Rightarrow\ \boxed{\frac{22^8}{(1+3+5+......+(2 \times 121-1))^2}} \\ \\ \\ \Rightarrow\ \boxed{\frac{22^8}{(121^2)^2}}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \boxed{\because\ 1+3+5+...+(2n-1)=n^2} \\ \\ \\ \Rightarrow\ \boxed{\frac{(11 \times 2)^8}{121^{2 \times 2}}} \\ \\ \\ \Rightarrow\ \boxed{\frac{11^8 \times 2^8}{121^4}} \\ \\ \\ \Rightarrow\ \boxed{\frac{11^8 \times 2^8}{(11^2)^4}} \\ \\ \\ \Rightarrow\ \boxed{\frac{22^8}{11^{2 \times 4}}}

\Rightarrow\ \boxed{\frac{11^8 \times 2^8}{11^8}} \\ \\ \\ \Rightarrow\ \boxed{2^8} \\ \\ \\ \Rightarrow\ \boxed{\bold{256}}

$$\sf{Plz ask me if you have any doubt on my answer. }

Thank\ you.\ $:-))$

   

Similar questions