Question

If m and n are +ve integers and m >n and if (1 + x)"+" (1 - x)"-" is expanded as a polynomial in x, then the coefficient of x2 is​

Answer 1

Answer:

2n

2

−m

This is a polynomial. Hence we differentiate it twice, the constant of the expression on differentiating twice, would be 2a

2

where a

2

is the coefficient of x

2

.

f(x)=(1+x)

m+n

(1−x)×m−n

f

′

(x)=(m+n)(1+x)

m+n−1

(1−x)×m−n−(m−n)(1+x)

m+n

(1−x)×m−n−1

f

′′

(x)=(m+n)(m+n−1)(1+x)

m+n−2

(1−x)×m−n−2(m+n)(m−n)(1+x)

m+n−1

(1−x)

×m−n−1+(m−n)(m−n−1)(1+x)

m+n

(1−x)×m−n−2

Putting x=0 to get constant,

Constant =2a

2

=(m+n)(m+n−1)−2(m+n)(m−n)+(m−n)(m−n−1)

Hence,

a

2

=(m+n)(m+n−1)−2(m+n)(m−n)+(m−n)(m−n−1)=2n

2

−m

Step-by-step explanation:

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