Класс 11 Решения RD Sharma – Глава 18 Биномиальная теорема – Упражнение 18.2 | Набор 3
Вопрос 27. Если 3- й , 4 -й , 5 -й и 6 -й члены разложения (x + α) n равны соответственно a, b, c и d, докажите, что 
.
Решение:
We are given, (x + α)n
So, T3 = a = nC2 xn-2 α2
T4 = b = nC3 xn-3 α3
T5 = c = nC4 xn-4 α4
T6 = d = nC5 xn-5 α5
We need to prove that,
=>
=>
=>
=>
=>
=>
=>
=>
=>
Hence proved.
Вопрос 28. Если 6 -й , 7 -й , 8 -й и 9 -й члены разложения (x + α) n равны соответственно a, b, c и d, докажите, что 
.
Решение:
We are given, (x + α)n
So, T6 = a = nC5 xn-5 α5
T7 = b = nC6 xn-6 α6
T8 = c = nC7 xn-7 α7
T9 = d = nC8 xn-8 α8
We need to prove that,
=>
=>
=>
=>
=>
=>
=>
=>
=>
Hence proved.
Вопрос 29. Если коэффициенты трех последовательных членов разложения (1+x) n равны соответственно 76, 95 и 76, найти n.
Решение:
We are given, (1+x)n
Let the three consecutive terms be rth, (r+1)th and (r+2)th.
We know the coefficient of rth term of a binomial expression is given by nCr-1.
Coefficient of rth term = nCr-1 = 76
Coefficient of (r+1)th term = nCr+1-1 = nCr = 95
Coefficient of (r+2)th term = nCr+2-1 = nCr+1 = 76
Now,
=>
=>
=> 5n − 5r = 4r + 4
=> 5n − 9r = 4 . . . . (1)
Also,
=>
=> 4n − 4r + 4 = 5r
=> 4n − r = −4 . . . . (2)
Subtracting (2) from (1), we get,
=> n = 4 + 4
=> n = 8
Therefore, the value of n is 8.
Вопрос 30. Если 6 -е , 7 -е и 8 -е в разложении (x + a) n равны соответственно 112, 7 и 1/4, найти x, a и n.
Решение:
We are given, (x + a)n
Also, T6 = nC5 xn-5 a5 = 112
T7 = nC6 xn-6 a6 = 7
T8 = nC7 xn-7 a7 = 1/4
Now,
=>
=>
=>
=>
=>
. . . . (1)
Also,
=>
=>
=>
=>
=>
. . . . (2)
From (1) and (2), we get,
=>
=>
=> 3n − 18 = 2n − 10
=> n = 8
Putting n = 8 in (2), we get,
=>
=>
=> x = 8a
Now, nC5 xn-5 a5 = 112
=> 8C5 x8-5 a5 = 112
=> 8C5 (8a)3 a5 = 112
=>
=> a8 =
=> a8 =
=> a = 1/2
So, x = 8 (1/2) = 4
Therefore, the value of x, a and n is 4, 1/2 and 8 respectively.
Вопрос 31. Если 2- й , 3- й , и 4 -й в разложении (x + a) n будет соответственно 240, 720 и 1080 соответственно, найти x, a и n.
Решение:
We are given, (x + a)n
Also, T2 = nC1 xn-1 a = 240
T3 = nC2 xn-2 a2 = 720
T4 = nC3 xn-3 a3 = 1080
Now,
=>
=>
=>
=>
. . . . (1)
Also,
=>
=>
=>
=>
. . . . (2)
From (1) and (2), we get,
=>
=> 12n − 24 = 9n − 9
=> 3n = 15
=> n = 5
Putting n = 5 in (2), we get,
=>
=>
=>
=>
Now, nC1 xn-1 a = 240
=> 5C1 x5-1 (3x/2) = 240
=> 5C1 x5 (3/2) = 240
=>
=>
=> x5 = 32
=> x5 = 25
=> x = 2
So, a = (3/2) (2) = 3
Therefore, the value of x, a and n is 2, 3 and 5 respectively.
Вопрос 32. Найдите a, b и n в разложении (a+b) n , если первые три члена равны 729, 7290 и 30375 соответственно.
Решение:
We are given, (a+b)n
Also, T1 = nC0 an = 729
T2 = nC1 an-1 b1 = 7290
T3 = nC2 an-2 b2 = 30375
Now,
=>
=>
. . . . (1)
Also,
=>
=>
=>
=>
. . . . (2)
From (1) and (2), we get,
=>
=> 30n − 30 = 25n
=> 5n = 30
=> n = 6
So, nC0 an = 729
=> a6 = 36
=> a = 3
Putting a = 3 in (2), we get,
=>
=>
=> b = 5
Therefore, the value of a, b and n is 3, 5 and 6 respectively.
Вопрос 33. Найдите а, если коэффициенты при х 2 и х 3 в разложении (3+ах) 9 равны.
Решение:
We have, (3+ax)9 = 9C0 39 + 9C1 38 (ax)1 + 9C2 37 (ax)2 + 9C3 36 (ax)3 + . . . .
Coefficient of x2 = 9C2 37 a2
Coefficient of x3 = 9C3 36 a3
According to the question, we have,
=> 9C2 37 a2 = 9C3 36 a3
=>
=> 81 = 63 a
=> a = 9/7
Therefore, the value of a is 9/7.
Вопрос 34. Найдите а, если коэффициенты при х и х 3 в разложении (2+ах) 4 равны.
Решение:
We have, (2+ax)4 = 4C0 24 + 4C1 23 (ax)1 + 4C2 22 (ax)2 + 4C3 2 (ax)3 + . . . .
Coefficient of x = 4C1 23 a
Coefficient of x3 = 4C3 2 a3
According to the question, we have,
=> 4C1 23 a = 4C3 2 a3
=> 4C3 23 a = 4C3 2 a3
=> 8a = 2a3
=> 2a (a − 4) = 0
=> a = 0 or a = 4
Therefore, the value of a is 0 or 4.
Вопрос 35. Если свободный от x член в разложении 
равно 405, найдите значение k.
Решение:
We have,
The general term of this expression will be,
Tr+1 =
=
If the term is independent of x , we must have,
=>
=> 10 − r − 4r = 0
=> 5r = 10
=> r = 2
Therefore, the required term is 3rd term.
So, we have,
=>
= 405
=>
= 405
=>
= 405
=>
= 405
=> 45k2 = 405
=> k2 = 9
=> k = 3
Therefore, the value of k is 3.
Вопрос 36. Найдите шестой член разложения (y 1/2 + x 1/3 ) n , если биномиальный коэффициент третьего члена с конца равен 45.
Решение:
We have, (y + x)n
The third term of the expansion from the end is (n + 1 − 3 + 1)th term = (n − 1)th term.
=> Tn-1 = Tn-2+1 = nCn-2 (y1/2)n-(n-2) (x1/3)n-2
The coefficient of this term is given, i.e., 45.
=> nCn-2 = 45
=> n (n − 1)/2 = 45
=> n (n − 1) = 90
=> n2 − n − 90 = 0
=> n2 − 10n + 9n − 90 = 0
=> n(n−10) + 9 (n−10) = 0
=> n = 10 or n = −9 (ignored)
So, the sixth term of the expansion is T6 = T5+1
= 10C10-5 (y1/2)10-(10-5) (x1/3)10-5
= 10C5 (y1/2)5 (x1/3)5
= 252 (y5/2) (x5/3)
Вопрос 37. Если p — действительное число и средний член разложения (p/2 + 2) 8 равен 1120, найдите p.
Решение:
We have, (p/2 + 2)8
Total number of terms is 8 + 1 = 9 (odd number).
The middle term is (9+1)/2 = 5th term.
Therefore, we get T5 = T4+1 = 1120
=> 8C4 (p/2)8-4 (2)4 = 1120
=> 70 (p/2)4 (2)4 = 1120
=> 70p4 = 1120
=> p4 = 16
=> p4 = 24
=> p = 2
Therefore, the value of p is 2.
Вопрос 38. Найдите n в биноме 
, если отношение 7-го члена от начала к концу равно 1/6.
Решение:
We have,
7th term from the beginning is
.
And 7th term from the end is
.
According to the question, we have,
=>
=>
=>
=>
=> (n − 12)/3 = −1
=> n − 12 = −3
=> n = 9
Therefore, the value of n is 9.
Вопрос 39. Если седьмой член от начала и конца биномиального разложения 
равны, найдите n.
Решение:
We have,
7th term from the beginning is
.
And 7th term from the end is
.
According to the question, we have,
=>
=>
=> n − 12 = 12 − n
=> 2n = 24
=> n = 12
Therefore, the value of n is 12.
