If 1- x 1 -x-x2 1-x-x2-x3 - 1-x-x2-x3-xn - a0-a1x-a2x2-a3x3-amxm then -r-0mar has the value equal to n-k - -Find k -

For the sum of the coefficients, we substitute x=1.Then we get nbsp; 2× 3× 4...× n× (n+1) =(n+1)! .Hence k=1 . ...

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Sum of Coefficients of All Terms

If 1× x 1 ...

Question

If $(1 \times x) (1 + x + x^{2}) (1 + x + x^{2} + x^{3}) . . . . . (1 + x + x^{2} + x^{3} + . . . . . . + x^{n}) \equiv$
$a_{0} + a_{1} x + a_{2} x^{2} + a_{3} x^{3} + . . . . . . . a_{m} x^{m}$ then $m \sum r = 0 a_{r}$ has the value equal to $(n + k)!$ .Find $k$ ?

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(1 + x) (1 + x + x^{2}) . . . . . (1 + x + . . . . . + x^{n}) = a_{0} + a_{1} x + a_{2} x^{2} + a_{3} x^{3} + . . . . .

a_{0} + a_{2} + a_{4} + . . . . .

(1 + x) (1 + x + x^{2}) (1 + x + x^{2} + x^{3}) . . . . . . (1 + x + x^{2} + . . . . + x^{30}) = a_{0} + a_{1} x + a_{2} x^{2} + . . . . . . . . + a_{465} x^{465}

a_{0} + a_{2} + a_{4} + . . . . . . . . + a_{464}

(1 + x) (1 + x + x^{2}) . . . . . (1 + x + . . . . . + x^{n}) = a_{0} + a_{1} x + a_{2} x^{2} + a_{3} x^{3} + . . . . .

a_{0} + a_{2} + a_{4} + . . . . .

(1 + x + x^{2}) (1 - \frac{x}{1!} + \frac{x^{2}}{2!} - \frac{x^{3}}{3!} + \dots) = a_{0} + a_{1} x + a_{2} x^{2} + a_{3} x^{3} + a_{4} x^{4} + . . .

(1 + x) (1 + x + x^{2}) (1 + x + x^{2} + x^{3}) \dots + (1 + x + x^{2} + . . x^{n - 1}) = a_{o} + a_{1} x + a_{2} x^{2} + . . a_{m} m x^{m}

a_{0} + a_{1} + a_{2} + \dots \dots \dots + a_{m} =

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