Checkout JEE MAINS 2022 Question Paper Analysis : Checkout JEE MAINS 2022 Question Paper Analysis :

The total number of terms in the expansion of (x + y)100 + (x – y)100 after simplification is

1) 51

2) 202

3) 100

4) 50

Solution: (1) 51

If x and y are real numbers, then for all n belongs to N,

\(\begin{array}{l}(x+y)^{n}={ }^{n} C_{0} x^{n} y^{0}+{ }^{n} C_{1} x^{n-1} y^{1} \quad+{ }^{n} C_{2} x^{n-2} y^{2}+\ldots+{ }^{n} C_{r} x^{n-r} y^{r}+\ldots+{ }^{n} C_{n-1} x^{1} y^{n-1}+{ }^{n} C_{n} x^{0} y^{n}\\ \therefore(x+y)^{100}={ }^{100} C_{0} x^{100} y^{0}+{ }^{100} C_{1} x^{99} y+{ }^{100} C_{2} x^{98} y^{2}+{ }^{100} C_{3} x^{97} y^{3}+\ldots+{ }^{100} C_{100} x^{0} y^{100}\\ \quad \therefore(x+y)^{100}+(x-y)^{100} =\left[\left(x^{100}+{ }^{100} C_{1} x^{99} y+{ }^{100} C_{2} x^{98} y^{2}+ { }^{100} C_{3} x^{97} y^{3}\right.\right. \left.+\ldots+{ }^{100} C_{100} x^{0} y^{100}\right)\\ \text \ Similarly, \ (x-y)^{100}={ }^{100} C_{0} x^{100} y^{0} -{ }^{100} C_{1} x^{99}{ }^{10}+{ }^{100} C_{2} x^{98} y^{2} +\left(x^{100}-{ }^{100} C_{1} x^{99} y+{ }^{100} C_{2} x^{98} y^{2}-{ }^{100} C_{3} x^{97} y^{3}\right. \left.\ldots+{ }^{100} C_{100} y^{100}\right]\\ =2\left[x^{100}+{ }^{100} C_{2} x^{98} y^{2}+{ }^{100} C_{4} x^{96} y^{4}\right. \left.+\ldots+{ }^{100} C_{100} y^{100}\right]\\ =51\end{array} \)

Was this answer helpful?

 
   

2 (2)

(2)
(1)

Choose An Option That Best Describes Your Problem

Thank you. Your Feedback will Help us Serve you better.

Leave a Comment

Your Mobile number and Email id will not be published. Required fields are marked *

*

*

Ask
Question