Q63
jee_main_2025_04_april_evening
Properties of Binomial Coefficients
If 1^2 cdot left( ^15 mathrmC_1 right) + 2^2 cdot left( ^15 mathrmC_2 right) + 3^2 cdot left( ^15 mathrmC_3 right) + dots + 15^2 cdot left( ^15 mathrmC_15 right) = 2^mathrmm cdot 3^mathrmn cdot 5^mathrmk, where m, n, k ∈ N, then m + n + k is equal to :-
Solution
### Related Formula
The general property linking indices to binomial coefficients is:
r cdot binomnr = n cdot binomn-1r-1
### Core Logic
The given series can be structured using summation notation:
S = sum_r=1^15 r^2 cdot binom15r
Apply the identity r cdot binom15r = 15 cdot binom14r-1 to reduce one factor of r:
S = sum_r=1^15 r cdot left[ 15 cdot binom14r-1 right] = 15 sum_r=1^15 r cdot binom14r-1
### Step 1: Splitting the linear term
Rewrite the index variable r as (r - 1) + 1 to align with the binomial lower index:
S = 15 sum_r=1^15 big((r - 1) + 1big) cdot binom14r-1
S = 15 sum_r=1^15 (r - 1) cdot binom14r-1 + 15 sum_r=1^15 binom14r-1
Applying the property again to the first summation term: (r-1)binom14r-1 = 14binom13r-2:
S = 15 cdot 14 sum_r=2^15 binom13r-2 + 15 sum_r=1^15 binom14r-1
### Step 2: Evaluating the Sums and Prime Factorization
Using the standard total sum of binomial coefficients sum_k=0^n binomnk = 2^n:
S = 15 cdot 14 cdot 2^13 + 15 cdot 2^14
Factor out 15 cdot 2^13 from the expression:
S = 15 cdot 2^13 (14 + 2) = 15 cdot 2^13 (16) = 15 cdot 2^13 cdot 2^4
S = 15 cdot 2^17 = (3^1 cdot 5^1) cdot 2^17
Matching this with the given format 2^m cdot 3^n cdot 5^k, we identify: m = 17, n = 1, and k = 1.
### Step 3: Calculating the sum of exponents
Evaluating the targeted summation:
m + n + k = 17 + 1 + 1 = 19
### Pattern Recognition
For a series of the type sum r^2 binomnr, remember the standard identity shortcut: n(n-1)2^n-2 + n2^n-1. Plugging in n=15 instantly outputs 15(14)2^13 + 15(2^14), bypasses matching terms manually.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Mathematics: Binomial Theorem
