If (1 + x) (1 + x + x^2) (1 + x + x^2 + x^3) | vedclass

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(B) To find the sum of the coefficients $\sum_{r=0}^m a_r$,we substitute $x = 1$ into the given polynomial identity.Substituting $x = 1$ in the expres

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$(n + 1)!$

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(B) To find the sum of the coefficients $\sum_{r=0}^m a_r$,we substitute $x = 1$ into the given polynomial identity.Substituting $x = 1$ in the expression $(1 + x) (1 + x + x^2) (1 + x + x^2 + x^3) \dots (1 + x + x^2 + \dots + x^n)$:$= (1 + 1) (1 + 1 + 1^2) (1 + 1 + 1^2 + 1^3) \dots (1 + 1 + 1^2 + \dots + 1^n)$$= (2) (3) (4) \dots (n + 1)$$= 1 	imes 2 	imes 3 	imes 4 \dots 	imes (n + 1)$$= (n + 1)!$Thus,the sum of the coefficients is $(n + 1)!$.

If $(1 + x) (1 + x + x^2) (1 + x + x^2 + x^3) \dots (1 + x + x^2 + \dots + x^n) \equiv a_0 + a_1x + a_2x^2 + a_3x^3 + \dots + a_mx^m$,then $\sum_{r=0}^m a_r$ has the value equal to:

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