Animated Solution for Mathematics - Binomial Theorem: If the number of integral terms in the expansion of (33+55)n is exactly 33, then the least value of n is:
Select Answer:
Visualized Solution
The Binomial Expansion
Consider the expansion of (321+581)n.
We need to find the least value of n for exactly 33 integral terms.
The General Term Formula
The general term in the expansion of (a+b)n is:
Tr+1=(rn)an−rbr
Substituting Our Values
Substitute a=321 and b=581:
Tr+1=(rn)(321)n−r(581)r
Simplifying the Exponents
Multiply the powers using exponent rules:
Tr+1=(rn)32n−r58r
Condition for Integral Terms
For Tr+1 to be an integer, the fractional powers must be eliminated.
Condition 1: 8r must be an integer.
Condition 2: 2n−r must be an integer.
Analyzing the First Condition
8r∈Z⟹r must be a multiple of 8.
Possible values for r: 0,8,16,24,…
Analyzing the Second Condition
2n−r∈Z⟹n−r must be even.
Since r is a multiple of 8, r is even.
Therefore, n must also be even.
The Sequence of r
The valid values of r form an Arithmetic Progression (A.P.).
Sequence: 0,8,16,24,…
First term (a) = 0
Common difference (d) = 8
Finding the 33rd Term
We need exactly 33 integral terms.
Formula for the k-th term of an A.P.: rk=a+(k−1)d
Substitute k=33, a=0, d=8.
Calculating the Maximum r
r33=0+(33−1)×8
r33=32×8
r33=256
The Constraint on n
In binomial expansion, the index r cannot exceed n (r≤n).
To include the 33rd term (r=256), we must have n≥256.
The Least Value of n
The next valid term would be at r=264.
For exactly 33 terms, 256≤n<264.
The least possible value is n=256.
00:00 / 00:00
The Sigma Insight: General Term and Middle Term
Solution Diagram
Analyzing the Setup
We are tasked with finding the least value of n such that the expansion of (33+55)n contains exactly 33 integral terms.
Note that the expression provided in the prompt contains a slight discrepancy in the roots. Based on the logic of the problem, we define the expression as:
(33+55)n=(31/3+51/5)n
The Master Key
The General Term
In any binomial expansion (a+b)n, the general term is given by:
Tr+1=(rn)an−rbr
Substituting a=31/3 and b=51/5, the general term becomes:
Tr+1=(rn)(31/3)n−r(51/5)r=(rn)33n−r55r
The Gatekeepers
The Exponents
For a term to be an integer, the exponents of the prime bases 3 and 5 must be non-negative integers. This leads to two conditions:
1. The exponent of 5, 5r, must be an integer. This implies r must be a multiple of 5, i.e., r∈{0,5,10,…}.
2. The exponent of 3, 3n−r, must be an integer. This implies (n−r) must be a multiple of 3.
The Arithmetic Progression
Finding the Pattern
Since r must be a multiple of 5, we have r=5k for k=0,1,2,…. Substituting this into the second condition, we require:
3n−5k∈Z⟹n−5k=3m⟹n=3m+5k
For a fixed n, the values of r that yield integral terms are those where r is a multiple of 5 and r≡n(mod3). Since 5 and 3 are coprime, the values of r will form an arithmetic progression with a common difference of lcm(3,5)=15.
The valid values of r are r0,r0+15,r0+30,…,r0+15(N−1), where N is the number of terms. We are given N=33.
The Final Calculation
The 33rd term is given by r33=r0+(33−1)×15=r0+32×15=r0+480.
Since r must satisfy 0≤r≤n, the smallest possible value for the 33rd term occurs when r0 is the smallest non-negative integer satisfying the divisibility conditions. For n to be the least value, we set r0=0 (which implies n must be a multiple of 3).
Thus, the 33rd term index is r=480. To ensure this is the last term, we require n≥480.
The next potential integral term would occur at r=480+15=495. To ensure there are exactly 33 terms, we must have n<495.
The least value of n that satisfies the condition is 480.