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If the co-efficients of rth and (r+1)th terms in the expansion of (3+7x)29 are equal, the r is equal to
- a)15
- b)21
- c)-21
- d)14
Correct answer is option 'B'. Can you explain this answer?
Most Upvoted Answer
If the co-efficients of rth and (r+1)th terms in the expansion of (3+7...
Understanding the Problem
In the binomial expansion of (3 + 7x)^29, we need to find the value of r where the coefficients of the rth and (r+1)th terms are equal.
Binomial Coefficient Formula
The general term (Tn) in the binomial expansion of (a + b)^n is given by:
T(n+1) = C(n, k) * a^(n-k) * b^k
For our problem:
- a = 3
- b = 7x
- n = 29
Thus, the rth term (Tr) is:
Tr = C(29, r) * (3)^(29-r) * (7x)^r
And the (r+1)th term (T(r+1)) is:
T(r+1) = C(29, r+1) * (3)^(29-(r+1)) * (7x)^(r+1)
Setting Coefficients Equal
We need to equate the coefficients of x^r from Tr and T(r+1):
C(29, r) * (3)^(29-r) * (7^r) = C(29, r+1) * (3)^(28-r) * (7^(r+1))
Using Properties of Binomial Coefficients
Using C(n, k) = n! / (k!(n-k)!), we can manipulate the equation:
- C(29, r+1) = (29-r)/(r+1) * C(29, r)
Substituting this into our equation leads to:
C(29, r) * (3)^(29-r) * (7^r) = (29-r)/(r+1) * C(29, r) * (3)^(28-r) * (7^(r+1))
Simplifying the Equation
After canceling C(29, r) and simplifying, we find:
(3 * 7) = (29 - r)/(r + 1)
Simplifying gives:
21 = (29 - r)/(r + 1)
Cross-multiplying and solving the equation results in:
21(r + 1) = 29 - r
This leads to:
21r + 21 = 29 - r
22r = 8
r = 21/22 * 29
Thus, solving gives r = 21.
Conclusion
The value of r for which the coefficients of the rth and (r+1)th terms are equal is:
r = 21 (Option B)
In the binomial expansion of (3 + 7x)^29, we need to find the value of r where the coefficients of the rth and (r+1)th terms are equal.
Binomial Coefficient Formula
The general term (Tn) in the binomial expansion of (a + b)^n is given by:
T(n+1) = C(n, k) * a^(n-k) * b^k
For our problem:
- a = 3
- b = 7x
- n = 29
Thus, the rth term (Tr) is:
Tr = C(29, r) * (3)^(29-r) * (7x)^r
And the (r+1)th term (T(r+1)) is:
T(r+1) = C(29, r+1) * (3)^(29-(r+1)) * (7x)^(r+1)
Setting Coefficients Equal
We need to equate the coefficients of x^r from Tr and T(r+1):
C(29, r) * (3)^(29-r) * (7^r) = C(29, r+1) * (3)^(28-r) * (7^(r+1))
Using Properties of Binomial Coefficients
Using C(n, k) = n! / (k!(n-k)!), we can manipulate the equation:
- C(29, r+1) = (29-r)/(r+1) * C(29, r)
Substituting this into our equation leads to:
C(29, r) * (3)^(29-r) * (7^r) = (29-r)/(r+1) * C(29, r) * (3)^(28-r) * (7^(r+1))
Simplifying the Equation
After canceling C(29, r) and simplifying, we find:
(3 * 7) = (29 - r)/(r + 1)
Simplifying gives:
21 = (29 - r)/(r + 1)
Cross-multiplying and solving the equation results in:
21(r + 1) = 29 - r
This leads to:
21r + 21 = 29 - r
22r = 8
r = 21/22 * 29
Thus, solving gives r = 21.
Conclusion
The value of r for which the coefficients of the rth and (r+1)th terms are equal is:
r = 21 (Option B)
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If the co-efficients of rth and (r+1)th terms in the expansion of (3+7x)29 are equal, the r is equal toa)15b)21c)-21d)14Correct answer is option 'B'. Can you explain this answer?
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