Question
Question: Let m be the smallest positive integer such that the coefficient of \({{\text{x}}^{\text{2}}}\) in t...
Let m be the smallest positive integer such that the coefficient of x2 in the expansion of (1 + x)2 + (1 + x)3 + (1 + x)4 + ........... + (1 + x)49 + (1 + x)50 is (3n+1)51C3 for some positive integer n. Then the value of n is-
Solution
Hint-To solve this question, we need to know the basics of Binomial Theorem. i.e.(x+y)n = nΣr=0 nCr xn – r · yr. where,nCr = [r!(n - r)!]n!
Complete step-by-step answer:
And by using the above statement we will get the value of x2 in the given expansion.
Now, from above equation, we observe that,
(1 + x)2 →coefficient of x2= 2C2
(1 + x)3 →coefficient of x2= 3C2
(1 + x)4 →coefficient of x2= 4C2
Similarly, we say,
(1 + x)n →coefficient of x2= nC2
According to question:
(1 + x)2 + (1 + x)3 + (1 + x)4 + ........... + (1 + x)49 + (1 + x)50 →coefficient of x2= (3n+1)51C3
coefficient of x2 in given expression-
2C2+3C2+4C2+………….+49C2+50C2 m2
As we know,
nCr + nCr - 1 = n + 1Cr
3C3+3C2+4C2+…………+49C2+50C2 m2
4C3+4C2+5C2+…………+49C2+50C2 m2
5C3+5C2+………
6C3+6C2+7C2……
Similarly, by multiple simplification, we get,
49C3+49C2+50C2 m2
50C3+ (50C2 m2-50C2)+50C2
51C3+50C2(m2-1)
Now, according to question-
51C3+50C2(m2-1) = (3n+1)51C3
51C3+50C2(m2-1) = 3n51C3+51C3
50C2(m2-1) = 3n[3!(51 - 3)!]51!
3n×351 [2!(51 - 2)!]50!= 50C2(m2-1)
3n×351 × 50C2= 50C2(m2-1)
51n+1=m2
Since, m be the smallest positive integer for some positive integer n.
And select n in such a way that m is a perfect square.
So, if we take n=5
We get m2= 256, which is a perfect square.
Thus, the value of n will be 5 in this question.
Note- The total number of terms in the expansion of (x+y)n are (n+1). The sum of exponents of x and y is always n. and The binomial coefficients which are equidistant from the beginning and from the ending are equal i.e. nC0 = nCn, nC1 = nCn-1 , nC2 = nCn-2 ,….. etc.