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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{{\text{x}}^{\text{2}}} in the expansion of (1 + x)2 + (1 + x)3 + (1 + x)4 + ........... + (1 + x)49 + (1 + x)50{\left( {{\text{1 + x}}} \right)^{\text{2}}}{\text{ + }}{\left( {{\text{1 + x}}} \right)^{\text{3}}}{\text{ + }}{\left( {{\text{1 + x}}} \right)^{\text{4}}}{\text{ + }}...........{\text{ + }}{\left( {{\text{1 + x}}} \right)^{{\text{49}}}}{\text{ + }}{\left( {{\text{1 + x}}} \right)^{{\text{50}}}} is (3n+1)51C3{}^{{\text{51}}}{{\text{C}}_{\text{3}}} for some positive integer n. Then the value of n is-

Explanation

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 = n![r!(n - r)!]{}^{\text{n}}{{\text{C}}_{\text{r}}}{\text{ = }}\dfrac{{{\text{n}}!}}{{{\text{[r}}!{\text{(n - r)}}!{\text{]}}}}

Complete step-by-step answer:
And by using the above statement we will get the value of x2{{\text{x}}^{\text{2}}} in the given expansion.
Now, from above equation, we observe that,
(1 + x)2{\left( {{\text{1 + x}}} \right)^{\text{2}}} \tocoefficient of x2{{\text{x}}^{\text{2}}}= 2C2{}^{\text{2}}{{\text{C}}_{\text{2}}}
(1 + x)3{\left( {{\text{1 + x}}} \right)^3} \tocoefficient of x2{{\text{x}}^{\text{2}}}= 3C2{}^3{{\text{C}}_{\text{2}}}
(1 + x)4{\left( {{\text{1 + x}}} \right)^4} \tocoefficient of x2{{\text{x}}^{\text{2}}}= 4C2{}^4{{\text{C}}_{\text{2}}}
Similarly, we say,
(1 + x)n{\left( {{\text{1 + x}}} \right)^{\text{n}}} \tocoefficient of x2{{\text{x}}^{\text{2}}}= nC2{}^n{{\text{C}}_{\text{2}}}
According to question:
(1 + x)2 + (1 + x)3 + (1 + x)4 + ........... + (1 + x)49 + (1 + x)50{\left( {{\text{1 + x}}} \right)^{\text{2}}}{\text{ + }}{\left( {{\text{1 + x}}} \right)^{\text{3}}}{\text{ + }}{\left( {{\text{1 + x}}} \right)^{\text{4}}}{\text{ + }}...........{\text{ + }}{\left( {{\text{1 + x}}} \right)^{{\text{49}}}}{\text{ + }}{\left( {{\text{1 + x}}} \right)^{{\text{50}}}} \tocoefficient of x2{{\text{x}}^{\text{2}}}= (3n+1)51C3{}^{{\text{51}}}{{\text{C}}_{\text{3}}}
coefficient of x2{{\text{x}}^{\text{2}}} in given expression-
2C2{}^{\text{2}}{{\text{C}}_{\text{2}}}+3C2{}^3{{\text{C}}_{\text{2}}}+4C2{}^4{{\text{C}}_{\text{2}}}+………….+49C2{}^{49}{{\text{C}}_{\text{2}}}+50C2{}^{50}{{\text{C}}_{\text{2}}} m2{{\text{m}}^{\text{2}}}
As we know,
nCr + nCr - 1 = n + 1Cr{}^{\text{n}}{{\text{C}}_{\text{r}}}{\text{ + }}{}^{\text{n}}{{\text{C}}_{{\text{r - 1}}}}{\text{ = }}{}^{{\text{n + 1}}}{{\text{C}}_{\text{r}}}
3C3{}^{\text{3}}{{\text{C}}_3}+3C2{}^{\text{3}}{{\text{C}}_{\text{2}}}+4C2{}^4{{\text{C}}_{\text{2}}}+…………+49C2{}^{49}{{\text{C}}_{\text{2}}}+50C2{}^{50}{{\text{C}}_{\text{2}}} m2{{\text{m}}^{\text{2}}}
4C3{}^4{{\text{C}}_3}+4C2{}^4{{\text{C}}_{\text{2}}}+5C2{}^5{{\text{C}}_{\text{2}}}+…………+49C2{}^{49}{{\text{C}}_{\text{2}}}+50C2{}^{50}{{\text{C}}_{\text{2}}} m2{{\text{m}}^{\text{2}}}
5C3{}^5{{\text{C}}_3}+5C2{}^5{{\text{C}}_{\text{2}}}+………
6C3{}^6{{\text{C}}_3}+6C2{}^6{{\text{C}}_{\text{2}}}+7C2{}^7{{\text{C}}_2}……
Similarly, by multiple simplification, we get,
49C3{}^{49}{{\text{C}}_3}+49C2{}^{49}{{\text{C}}_{\text{2}}}+50C2{}^{50}{{\text{C}}_{\text{2}}} m2{{\text{m}}^{\text{2}}}
50C3{}^{50}{{\text{C}}_3}+ (50C2{}^{50}{{\text{C}}_{\text{2}}} m2{{\text{m}}^{\text{2}}}-50C2{}^{50}{{\text{C}}_{\text{2}}})+50C2{}^{50}{{\text{C}}_{\text{2}}}
51C3{}^{51}{{\text{C}}_3}+50C2{}^{50}{{\text{C}}_{\text{2}}}(m2{{\text{m}}^{\text{2}}}-1)
Now, according to question-
51C3{}^{51}{{\text{C}}_3}+50C2{}^{50}{{\text{C}}_{\text{2}}}(m2{{\text{m}}^{\text{2}}}-1) = (3n+1)51C3{}^{{\text{51}}}{{\text{C}}_{\text{3}}}
51C3{}^{51}{{\text{C}}_3}+50C2{}^{50}{{\text{C}}_{\text{2}}}(m2{{\text{m}}^{\text{2}}}-1) = 3n51C3{}^{51}{{\text{C}}_3}+51C3{}^{51}{{\text{C}}_3}
50C2{}^{50}{{\text{C}}_{\text{2}}}(m2{{\text{m}}^{\text{2}}}-1) = 3n51![3!(51 - 3)!]\dfrac{{51!}}{{{\text{[3}}!{\text{(51 - 3)}}!{\text{]}}}}
3n×513 \times \dfrac{{51}}{3} 50![2!(51 - 2)!]\dfrac{{50!}}{{{\text{[2}}!{\text{(51 - 2)}}!{\text{]}}}}= 50C2{}^{50}{{\text{C}}_{\text{2}}}(m2{{\text{m}}^{\text{2}}}-1)

3n×513 \times \dfrac{{51}}{3} ×\times 50C2{}^{50}{{\text{C}}_2}= 50C2{}^{50}{{\text{C}}_{\text{2}}}(m2{{\text{m}}^{\text{2}}}-1)
51n+1=m2{{\text{m}}^{\text{2}}}
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{{\text{m}}^{\text{2}}}= 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.