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P2 Binomial - Problem.
Question from Heinamann P2 Book: p48 Q12 :
The coefficients of the x and x2 terms in the expansion of (1+kx)n are 44 and 924 respectivley. Find the values of the constants k and n.
(Answer in back: n = 22, k = 2)
Can someone explain to me how to do this? The thing is, i arrived at the answer but i dont know why i did because when i carried out the expansion and equated the coefficients with their respective expansions, i acciently left out the 2! value for the x2, but somehow arrived at the answer.
When i re-tried putting in the missing 2!, i did not get the correct answer.
Could someone help me out!
The coefficients of the x and x2 terms in the expansion of (1+kx)n are 44 and 924 respectivley. Find the values of the constants k and n.
(Answer in back: n = 22, k = 2)
Can someone explain to me how to do this? The thing is, i arrived at the answer but i dont know why i did because when i carried out the expansion and equated the coefficients with their respective expansions, i acciently left out the 2! value for the x2, but somehow arrived at the answer.
When i re-tried putting in the missing 2!, i did not get the correct answer.
Could someone help me out!
cherc2005
Question from Heinamann P2 Book: p48 Q12 :
The coefficients of the x and x2 terms in the expansion of (1+kx)n are 44 and 924 respectivley. Find the values of the constants k and n.
(Answer in back: n = 22, k = 2)
The coefficients of the x and x2 terms in the expansion of (1+kx)n are 44 and 924 respectivley. Find the values of the constants k and n.
(Answer in back: n = 22, k = 2)
(1 + kx)n
= 1 + nkx + [n(n-1)/2!][kx]2 + ....
nk = 44 => k = 44/n
[n(n-1)/2][k²] = 924
[½n² - ½n]k² = 924
[½n² - ½n][44/n]² = 924
[½n² - ½n][1936/n²] = 924
1936/2 - 1936/2n = 924
1936n - 1936 = 1848n
88n = 1936
n = 1936/88 = 22
k = 44/n = 44/22 = 2
cherc2005
Question from Heinamann P2 Book: p48 Q12 :
The coefficients of the x and x2 terms in the expansion of (1+kx)n are 44 and 924 respectivley. Find the values of the constants k and n.
(Answer in back: n = 22, k = 2)
Can someone explain to me how to do this? The thing is, i arrived at the answer but i dont know why i did because when i carried out the expansion and equated the coefficients with their respective expansions, i acciently left out the 2! value for the x2, but somehow arrived at the answer.
When i re-tried putting in the missing 2!, i did not get the correct answer.
Could someone help me out!
The coefficients of the x and x2 terms in the expansion of (1+kx)n are 44 and 924 respectivley. Find the values of the constants k and n.
(Answer in back: n = 22, k = 2)
Can someone explain to me how to do this? The thing is, i arrived at the answer but i dont know why i did because when i carried out the expansion and equated the coefficients with their respective expansions, i acciently left out the 2! value for the x2, but somehow arrived at the answer.
When i re-tried putting in the missing 2!, i did not get the correct answer.
Could someone help me out!
1 + nkx + n(n-1)/2 k^2 x^2 + ... = 1 + 44 x + 924 x^2 + ...
So
nk = 44
n(n-1)k^2 = 2 x 924 = 1848
So
n(n-1)/n^2 = 1848/44^2 = 21/22 [cancelling down]
1 - 1/n = 21/22
n = 22
k = 2
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