If binomial coefficients of three consecutive terms of (1 + x)n are in...
Explanation:
To find the maximum value of n, we need to consider the binomial coefficients of three consecutive terms of (1 + x)^n in harmonic progression (HP).
Let's first write the expansion of (1 + x)^n using the binomial theorem:
(1 + x)^n = C(n, 0) + C(n, 1)x + C(n, 2)x^2 + ... + C(n, n-1)x^(n-1) + C(n, n)x^n
Where C(n, k) represents the binomial coefficient of n choose k.
Binomial Coefficients in Harmonic Progression:
In a harmonic progression (HP), the reciprocals of the terms are in arithmetic progression (AP). Let's consider the reciprocals of the binomial coefficients:
1/C(n, 0), 1/C(n, 1), 1/C(n, 2), ..., 1/C(n, n-1), 1/C(n, n)
We need to check if these reciprocals are in arithmetic progression.
Reciprocal of a Binomial Coefficient:
The reciprocal of a binomial coefficient C(n, k) is given by:
1/C(n, k) = (n!)/((n-k)!k!)
Reciprocal of Binomial Coefficients in HP:
Let's calculate the reciprocals of the binomial coefficients:
1/C(n, 0) = 1/1 = 1
1/C(n, 1) = 1/n
1/C(n, 2) = 2!/(n-2)!(2!)
= 2/(n(n-1))
1/C(n, 3) = 3!/(n-3)!(3!)
= 6/((n)(n-1)(n-2))
Similarly, we can calculate the reciprocals of the remaining binomial coefficients.
Checking for Arithmetic Progression:
To check if the reciprocals of the binomial coefficients are in arithmetic progression, we need to check if the differences between consecutive terms are constant.
Let's calculate the differences between consecutive terms:
1/n - 1 = (1 - n)/n
2/(n(n-1)) - 1/n = (2 - n)/(n(n-1))
6/((n)(n-1)(n-2)) - 2/(n(n-1)) = (6(n-2) - 2(n)(n-1))/((n)(n-1)(n-2))
...
We can see that the differences between consecutive terms are not constant. Therefore, the reciprocals of the binomial coefficients are not in arithmetic progression.
Conclusion:
Since the reciprocals of the binomial coefficients are not in arithmetic progression, the binomial coefficients themselves are not in harmonic progression.
Hence, there is no maximum value of n for which the binomial coefficients of three consecutive terms of (1 + x)^n are in harmonic progression.
Therefore, the correct answer is option 'D' (None of these).
If binomial coefficients of three consecutive terms of (1 + x)n are in...
Let the coefficients of rth, (r + 1)th, and (r + 2)th terms be in HP.
Then,
⇒ n
2 – 4nr + 4r
2 + n = 0
⇒ (n – 2r)
2 + n = 0
which is not possible for any value for n.