Without using binomial or any -nomial theorem and with logical reasoning find out the co-efficient of the following in the expansion of
(x+1)(x+2)(x+3)...............(x+n) :
1) xn-1
2) x
Ans :
1) 1+2+3+......+n
2) n!(1+12+13+...............+1n)
1coefficient of xn-1 can be obtained by taking a constant out from any one of the factors in all possible ways and then combining with the 1(i.e xn-1 coefficient ,all others will be less than xn-1) in left..
i.e
if we choose (x+1)
1 is the constant
that on combining with xn-1 coeffivcient i.e gives xn-1 coefficient
similarly we may choose x+2
constant here is 2
similarly if we procced
we will get xn-1 coefficient as
(1+2+3+..........+n)
62aveek if you are not able to understand this after karna's post,
Try to multiply (x+1)(x+2)(x+3) completely term by term so that you have 8 terms and then try to find the coeff of x^2
341Let
(x+1)(x+2)...(x+n) = x^n+a_{n-1}x^{n-1}+...+a_0 = P(x)
Then the roots of P(x) are -1,-2,..,-n
Then by Vieta's relations,
an-1 = sum of roots = 1+2+3+...+n
Now if you consider the reciprocal polynomial of P(x) i.e.
Q(x) = n!x^n+a_1x^{n-1}+...+1 its roots are
-1, -\frac{1}{2}, -\frac{1}{3},...,-\frac{1}{n}
and so again by Vieta's relations a_1 = n! \left(1 + \frac{1}{2}+...+\frac{1}{n} \right)
