The coefficient of x^m in (1+x)^p +(1+x)^(p+1) +...+(1+x)^n , p≤ m≤ n is


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Question Number 62086 by Cypher1207 last updated on 15/Jun/19

$The coefficient of x^{m} in$ ${(1 + x)}^{p} + {(1 + x)}^{p + 1} + . . . + {(1 + x)}^{n}, p ⩽ m ⩽ n$ $is$
	Answered by mr W last updated on 15/Jun/19
	$(1+x)^p +(1+x)^(p+1) +...+(1+x)^n ← GP =(((1+x)^p [(1+x)^(n−p+1) −1])/((1+x)−1)) =(((1+x)^(n+1) −(1+x)^p )/x) =(1/x)[Σ_(k=0) ^(n+1) C_k ^(n+1) x^k −Σ_(k=0) ^p C_k ^p x^k ] =(1/x)[Σ_(k=0) ^p C_k ^(n+1) x^k −Σ_(k=0) ^p C_k ^p x^k +Σ_(k=p+1) ^(n+1) C_k ^(n+1) x^k ] =(1/x)[Σ_(k=0) ^p (C_k ^(n+1) −C_k ^p )x^k +Σ_(k=p+1) ^(n+1) C_k ^(n+1) x^k ] =Σ_(k=0) ^p (C_k ^(n+1) −C_k ^p )x^(k−1) +Σ_(k=p+1) ^(n+1) C_k ^(n+1) x^(k−1) =Σ_(k=0) ^(p−1) (C_(k+1) ^(n+1) −C_(k+1) ^p )x^k +Σ_(k=p) ^n C_(k+1) ^(n+1) x^k =Σ_(k=0) ^n a_k x^k with a_k = { ((C_(k+1) ^(n+1) −C_(k+1) ^p for 0≤k≤p−1)),((C_(k+1) ^(n+1) for p≤k≤n)) :} for p≤m≤n, coef. of x^m is C_(m+1) ^(n+1) .$
	${(1 + x)}^{p} + {(1 + x)}^{p + 1} + . . . + {(1 + x)}^{n} \leftarrow G P$ $= \frac{{(1 + x)}^{p} [{(1 + x)}^{n - p + 1} - 1]}{(1 + x) - 1}$ $= \frac{{(1 + x)}^{n + 1} - {(1 + x)}^{p}}{x}$ $= \frac{1}{x} [\underset{k = 0}{\sum^{n + 1}} C_{k}^{n + 1} x^{k} - \underset{k = 0}{\sum^{p}} C_{k}^{p} x^{k}]$ $= \frac{1}{x} [\underset{k = 0}{\sum^{p}} C_{k}^{n + 1} x^{k} - \underset{k = 0}{\sum^{p}} C_{k}^{p} x^{k} + \underset{k = p + 1}{\sum^{n + 1}} C_{k}^{n + 1} x^{k}]$ $= \frac{1}{x} [\underset{k = 0}{\sum^{p}} (C_{k}^{n + 1} - C_{k}^{p}) x^{k} + \underset{k = p + 1}{\sum^{n + 1}} C_{k}^{n + 1} x^{k}]$ $= \underset{k = 0}{\sum^{p}} (C_{k}^{n + 1} - C_{k}^{p}) x^{k - 1} + \underset{k = p + 1}{\sum^{n + 1}} C_{k}^{n + 1} x^{k - 1}$ $= \underset{k = 0}{\sum^{p - 1}} (C_{k + 1}^{n + 1} - C_{k + 1}^{p}) x^{k} + \underset{k = p}{\sum^{n}} C_{k + 1}^{n + 1} x^{k}$ $= \underset{k = 0}{\sum^{n}} a_{k} x^{k}$ $w i t h a_{k} = {\begin{cases} C_{k + 1}^{n + 1} - C_{k + 1}^{p} f o r 0 ⩽ k ⩽ p - 1 \\ C_{k + 1}^{n + 1} f o r p ⩽ k ⩽ n \end{cases}$ $f o r p ⩽ m ⩽ n,$ $c o e f . o f x^{m} i s C_{m + 1}^{n + 1} .$
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