Find the largest co-efficient in the expansion of (1 + x)n, given that...
We know that, the coefficients in a binomial expansion is obtained by replacing each variable by unit in the given expression.
Therefore, sum of the coefficients in (a+b)^n
= 4096=(1+1)n
⇒ 4096=(2)n
⇒ (2)12=(2)n
⇒ n=12
Here n is even, so the greatest coefficient is nCn/2 i.e., 12C6 = 924
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Find the largest co-efficient in the expansion of (1 + x)n, given that...
Given:
The expansion of (1 + x)^n has a sum of coefficients equal to 4096.
To find:
The largest coefficient in the expansion of (1 + x)^n.
Solution:
Step 1: Expansion of (1 + x)^n
The expansion of (1 + x)^n can be written as:
(1 + x)^n = C(n, 0) + C(n, 1)x + C(n, 2)x^2 + ... + C(n, n)x^n
Where C(n, r) represents the binomial coefficient, which is given by the formula:
C(n, r) = n! / (r!(n-r)!)
Step 2: Sum of coefficients
The sum of coefficients in the expansion of (1 + x)^n is equal to the value of the expression when x is replaced by 1, i.e., x = 1. Therefore, substituting x = 1 in the expansion, we get:
(1 + 1)^n = C(n, 0) + C(n, 1) + C(n, 2) + ... + C(n, n)
Simplifying this expression gives us:
2^n = C(n, 0) + C(n, 1) + C(n, 2) + ... + C(n, n)
Given that the sum of coefficients is 4096, we can write the equation as:
2^n = 4096
Step 3: Solving for n
To find the value of n, we need to solve the equation 2^n = 4096. Taking the logarithm of both sides, we get:
log2(2^n) = log2(4096)
n log2(2) = log2(4096)
n = log2(4096) / log2(2)
n = 12
Therefore, the value of n is 12.
Step 4: Finding the largest coefficient
The largest coefficient in the expansion of (1 + x)^n occurs when x^n is the term with the largest power of x. In this case, the largest power of x is x^6, which occurs when r = 6 in the binomial coefficient C(n, r).
Therefore, the largest coefficient is C(12, 6) or 12C6.
Hence, the correct answer is 12C6.