Q65
jee_main_2025_02_april_evening
Properties of Binomial Coefficients
If sum_mathrmr = 0^10left(frac10^mathrmr + 1 - 110^mathrmrright) cdot binom11r + 1 = fracalpha^11 - 11^1110^10, then alpha is equal to:
Solution
### Related Formula
textSum of binomial coefficients: sum_k=0^n binomnk x^k = (1 + x)^n
textShifted binomial sum: sum_k=1^n binomnk x^k = (1+x)^n - 1
### Core Logic
We split the summation into two parts, express each as a binomial series expansion, and equate the resulting algebraic fraction to solve for alpha.
### Step 1: Split the summation
The general term inside the summation can be written as:
left( frac10^r+1 - 110^r right) = 10 - frac110^r = 10 - 10 left( frac110 right)^r+1
Substitute this back into the sum:
S = sum_r=0^10 left[ 10 - 10 left( frac110 right)^r+1 right] binom11r+1
S = 10 sum_r=0^10 binom11r+1 - 10 sum_r=0^10 binom11r+1 left( frac110 right)^r+1
### Step 2: Evaluate both parts of the sum
For the first part, let s = r+1:
sum_r=0^10 binom11r+1 = sum_s=1^11 binom11s = 2^11 - 1
For the second part, using s = r+1:
sum_r=0^10 binom11r+1 left( frac110 right)^r+1 = sum_s=1^11 binom11s left( frac110 right)^s = left( 1 + frac110 right)^11 - 1 = left( frac1110 right)^11 - 1
### Step 3: Combine and find alpha
Multiply both parts by 10:
S = 10 left( 2^11 - 1 right) - 10 left[ left( frac1110 right)^11 - 1 right]
S = 10 cdot 2^11 - 10 - 10 cdot frac11^1110^11 + 10 = 10 cdot 2^11 - frac11^1110^10
Express the first term with a denominator of 10^10:
10 cdot 2^11 = frac10^11 cdot 2^1110^10 = frac20^1110^10
Thus, the total sum is:
S = frac20^11 - 11^1110^10
Comparing this with fracalpha^11 - 11^1110^10, we find:
alpha = 20
### Pattern Recognition
Binomial base scaling: Whenever you see a sum of the form sum a^r binomnr, it is simply the expanded form of a shifted binomial expansion of (1 + a)^n. Factoring out scaling constants yields standard analytical forms.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Mathematics: Binomial Theorem
