find tbe polynom p withdegre 5 wich verify p(x+1)−p(x)=x^4 and p(0)=0 for that put p(x)=ax^5 +bx^4 +cx^3 +dx^2 +ex +f and find the coefficients. 2) find interms of n the value of sum 1 +2^4 +3^4 +....+n^4 .


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Question Number 38740 by abdo.msup.com last updated on 29/Jun/18

$f i n d t b e p o l y n o m p w i t h d e g r e 5 w i c h v e r i f y$ $p (x + 1) - p (x) = x^{4} a n d p (0) = 0$ $f o r t h a t p u t p (x) = {a x}^{5} + {b x}^{4} + {c x}^{3} + {d x}^{2}$ $+ e x + f a n d f i n d t h e c o e f f i c i e n t s .$ $2) f i n d i n t e r m s o f n t h e v a l u e o f$ $s u m 1 + 2^{4} + 3^{4} + . . . . + n^{4} .$
Commented by prakash jain last updated on 29/Jun/18

${(n + 1)}^{5} - n^{5} = 5 n^{4} + 10 n^{3} + 10 n^{2} + 5 n + 1$ $\underset{i = 0}{\sum^{n}} {(i + 1)}^{5} - i^{5} = \underset{i = 1}{\sum^{n}} 5 i^{4} + 10 i^{3} + 10 i^{2} + 5 i + 1$ $\underset{i = 1}{\sum^{n}} i^{4} = \frac{1}{5} {(n + 1)}^{5} - \frac{n^{2} {(n + 1)}^{2}}{2}$ $- \frac{n (n + 1) (2 n + 1)}{3} - \frac{n (n + 1)}{2} - \frac{n}{5}$
Commented by math khazana by abdo last updated on 29/Jun/18

$s i r P r a k a s h a r e y o u s t a y i n g i n i n d i a o r a b r o a d . .$
Commented by prakash jain last updated on 30/Jun/18

$I am in India .$
Commented by math khazana by abdo last updated on 30/Jun/18

$g o o d s i r i h o p e t o t a k e a t r i p t o i n d i a a n d s e e$ $t h e h i m a l a y a m o u n t a i n s a n d o t h e r p l a c e s . . .$
	Answered by MrW3 last updated on 29/Jun/18

	$p (x) = {a x}^{5} + {b x}^{4} + {c x}^{3} + {d x}^{2} + e x + f$ $p (0) = 0 \Rightarrow f = 0$ $p (x + 1) = a {(x + 1)}^{5} + b {(x + 1)}^{4} + c {(x + 1)}^{3} + d {(x + 1)}^{2} + e (x + 1)$ $p (x + 1) - p (x) = a [{(x + 1)}^{5} - x^{5}] + b [{(x + 1)}^{4} - x^{4}] + c [{(x + 1)}^{3} - x^{3}] + d [{(x + 1)}^{2} - x^{2}] + e = x^{4}$ $a [5 x^{4} + 10 x^{3} + 10 x^{2} + 5 x + 1] + b [4 x^{3} + 6 x^{2} + 4 x + 1] + c [3 x^{2} + 3 x + 1] + d [2 x + 1] + e = x^{4}$ $5 {a x}^{4} + (10 a + 4 b) x^{3} + (10 a + 6 b + 3 c) x^{2} + (5 a + 4 b + 3 c + 2 d) x + (a + b + c + d + e) = x^{4}$ $\Rightarrow 5 a = 1 \Rightarrow a = \frac{1}{5}$ $\Rightarrow 10 a + 4 b = 0 \Rightarrow b = - \frac{5}{2} a = - \frac{1}{2}$ $\Rightarrow 10 a + 6 b + 3 c = 0 \Rightarrow c = - \frac{10}{3} a - 2 b = - \frac{1}{6} + 1 = - \frac{5}{6}$ $\Rightarrow 5 a + 4 b + 3 c + 2 d = 0 \Rightarrow d = - \frac{5}{2} a - 2 b - \frac{3}{2} c = - \frac{1}{2} + 1 + \frac{5}{4} = \frac{7}{4}$ $\Rightarrow a + b + c + d + e = 0 \Rightarrow e = - a - b - c - d = - \frac{1}{5} + \frac{1}{2} + \frac{5}{6} - \frac{7}{4} = \frac{- 12 + 30 + 50 - 85}{60} = - \frac{17}{60}$
	Commented by math khazana by abdo last updated on 29/Jun/18

	$t h a n k y o u s i r .$
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