22. Show that
(i) [tex]$\binom{n}{0}+\binom{n}{1}+\binom{n}{2}+\cdots\binom{n}{n}=2^n$[/tex]
(ii) [tex]$\binom{n}{0}-\binom{n}{1}+\binom{n}{2}-\cdots-\binom{n}{n}=0$[/tex]
23. When [tex]$(1-2 x)^p$[/tex] is expanded, the coefficient of [tex]$x^2$[/tex] is 40. Given that [tex]$p\ \textgreater \ 0$[/tex], use this to find
(i) the value of the constant [tex]$p$[/tex]
(ii) the coefficient of [tex]$x$[/tex]
(iii) the coefficient of [tex]$x^2$[/tex]
Answer (2)
Use the binomial theorem with x = 1 and y = 1 to show ( 0 n ) + ( 1 n ) + ( 2 n ) + ⋯ + ( n n ) = 2 n .
Use the binomial theorem with x = 1 and y = − 1 to show ( 0 n ) − ( 1 n ) + ( 2 n ) − ⋯ + ( − ( n n ) ) = 0 .
Expand ( 1 − 2 x ) p and equate the coefficient of x 2 to 40 to find p = 5 .
Determine the coefficient of x as − 2 p = − 10 , so the final answers are p = 5 and − 10 .
Explanation
Introduction Let's tackle these problems step by step! We'll start with the binomial theorem and its applications.
The Binomial Theorem We'll use the binomial theorem, which states that for any non-negative integer n and any real numbers x and y :
( x + y ) n = k = 0 ∑ n ( k n ) x n − k y k This theorem is the foundation for expanding expressions of the form ( x + y ) n .
Proof of 22(i) For question 22(i), we want to show that ( 0 n ) + ( 1 n ) + ( 2 n ) + ⋯ + ( n n ) = 2 n . We can use the binomial theorem by setting x = 1 and y = 1 . Then we have: ( 1 + 1 ) n = k = 0 ∑ n ( k n ) 1 n − k 1 k = k = 0 ∑ n ( k n ) = ( 0 n ) + ( 1 n ) + ( 2 n ) + ⋯ + ( n n ) Since ( 1 + 1 ) n = 2 n , we have shown that ( 0 n ) + ( 1 n ) + ( 2 n ) + ⋯ + ( n n ) = 2 n .
Proof of 22(ii) For question 22(ii), we want to show that ( 0 n ) − ( 1 n ) + ( 2 n ) − ⋯ + ( − ( n n ) ) = 0 . Again, we use the binomial theorem, but this time we set x = 1 and y = − 1 . Then we have: ( 1 + ( − 1 ) ) n = k = 0 ∑ n ( k n ) 1 n − k ( − 1 ) k = ( 0 n ) − ( 1 n ) + ( 2 n ) − ⋯ + ( − ( n n ) ) Since ( 1 − 1 ) n = 0 n = 0 , we have shown that ( 0 n ) − ( 1 n ) + ( 2 n ) − ⋯ + ( − ( n n ) ) = 0 .
Expanding (1-2x)^p Now let's move on to question 23. We are given that the coefficient of x 2 in the expansion of ( 1 − 2 x ) p is 40, and we need to find the value of p , the coefficient of x , and verify the coefficient of x 2 .
Using the binomial theorem, we expand ( 1 − 2 x ) p :
( 1 − 2 x ) p = ( 0 p ) 1 p ( − 2 x ) 0 + ( 1 p ) 1 p − 1 ( − 2 x ) 1 + ( 2 p ) 1 p − 2 ( − 2 x ) 2 + ⋯ ( 1 − 2 x ) p = 1 + p ( − 2 x ) + 2 p ( p − 1 ) ( − 2 x ) 2 + ⋯ ( 1 − 2 x ) p = 1 − 2 p x + 2 p ( p − 1 ) x 2 + ⋯
Finding the value of p For question 23(i), we are given that the coefficient of x 2 is 40. From our expansion, the coefficient of x 2 is 2 p ( p − 1 ) . Therefore, we have the equation: 2 p ( p − 1 ) = 40 p ( p − 1 ) = 20 p 2 − p − 20 = 0 Factoring the quadratic, we get ( p − 5 ) ( p + 4 ) = 0 . The solutions are p = 5 and p = − 4 . Since we are given that 0"> p > 0 , we must have p = 5 .
Finding the coefficient of x For question 23(ii), we need to find the coefficient of x . From our expansion, the coefficient of x is − 2 p . Since we found that p = 5 , the coefficient of x is − 2 ( 5 ) = − 10 .
Verifying the coefficient of x^2 For question 23(iii), we are asked to find the coefficient of x 2 , but we already know it is 40, as given in the problem statement. We can verify this using our value of p = 5 . The coefficient of x 2 is 2 p ( p − 1 ) = 2 ( 5 ) ( 5 − 1 ) = 2 ( 5 ) ( 4 ) = 40 . This confirms the given information.
Final Answers In summary: (i) We have shown that ( 0 n ) + ( 1 n ) + ( 2 n ) + ⋯ + ( n n ) = 2 n .
(ii) We have shown that ( 0 n ) − ( 1 n ) + ( 2 n ) − ⋯ + ( − ( n n ) ) = 0 .
(i) The value of p is 5. (ii) The coefficient of x is -10. (iii) The coefficient of x 2 is 40.
Examples
The binomial theorem is used in probability calculations, such as determining the likelihood of getting a certain number of heads when flipping a coin multiple times. For example, if you flip a coin 10 times, the binomial theorem can help you calculate the probability of getting exactly 5 heads. This concept extends to various real-world scenarios, including genetics (predicting the traits of offspring), finance (modeling investment returns), and even sports (analyzing the probability of winning a game). Understanding binomial coefficients and expansions allows for precise calculations and predictions in these diverse fields.
We proved that the sums of binomial coefficients yield 2 to the power of n and zero for alternating signs. We calculated that p is 5, with the coefficient of x being -10, and confirmed the coefficient of x^2 as 40. This is consistent with the binomial theorem and its applications.
;
