Question

Algebra problem ,
prove the following is an identity:
x* ( 1 + x)^7/7 - (1 + x )^8/(7*8) = (1+x)^8/8 - (1+x)^7/7
You can start from the left side or right side.
(the problem arose when integrating x (1+x)^6 , i solved it two different ways using u sub and integration by parts).

Answer 1

according to wolfram it is a true equation
http://www.wolframalpha.com/input/?i=x*+%28+1+%2B+x%29^7%2F7+-+%281+%2B+x+%29^8%2F%287*8%29+%3D+%281%2Bx%29^8%2F8+-+%281%2Bx%29^7%2F7+

Answer 2

i made various attempts and got stuck

Answer 3

Attempt:
(1+x)^8/8 - (1+x)^7/7
= ( 1 + x)^7 [ ( 1 + x)/8 - 1/7 ]
= ( 1 + x)^7 [ ( 7 + 7x - 8 ) / (8*7) ]
= (1+x)^7 [ ( 7x - 1 ) / (8*7) ]
=. . .
= x* ( 1 + x)^7/7 - (1 + x )^8/(7*8)

Answer 4

left side:
\[=\frac{ 8x(1+x)^{7}-(1+x ^{8}) }{ 56 }=\frac{ (1+x)^{7}(7x-1) }{ 56 }\]
right side:
\[=\frac{ 7(1+x)^{8}-8(1+x)^{7} }{ 56 }=\frac{ (7x-1)(1+x)^{7} }{ 56 }\]

Answer 5

that works :)

Answer 6

you formed a bridge between right hand side and left hand side

Answer 7

yes! I solved your identity like a trigonometric identity!

Answer 8

so it is easier to show that both sides simplify to the same expression, and then go from there

Answer 9

yes!

Answer 10

didn't think of that :)

Answer 11

thanks!

Answer 12

I MEANT ON MESSAGE.....

Answer 13

plz message them to me

Answer 14

Subtract the two sides and show that it is zer0
\[-\frac{1}{8} (x+1)^8-\frac{(x+1)^8}{7\times 8}+\frac{1}{7} (x+1)^7+\frac{1}{7}
(x+1)^7 x=\\
-\frac{1}{7} (x+1)^8+\frac{1}{7} (x+1)^7+\frac{1}{7} x
(x+1)^7=\\ \frac{1}{7} (x+1)^7
\left(\frac{x}{7}-\frac{x}{7}+\frac{1}{7}-\frac{1}{7}\right)=0\]

Answer 15

yes that works too :)
I thought there was a simple way to factor the right hand side of the original equation so that it would turn into the left hand side

Answer 16

but this is an acceptable way to prove an identity, as well

Answer 17

\[\text {One Side} = \frac {1} {7} x (x + 1)^7 - \frac {1} {56}
(x + 1)^8 = \\\frac {8} {57} x (x + 1)^7 - \frac {1} {56}
(x + 1)^8 = \\ \frac {1} {56} (x + 1)^7 (8
x - x - 1) = \frac {1} {56} (x + 1)^7 (7 x - 1)\]

Answer 18

\[\text {Other Side} = \frac {1} {8} (x + 1)^8 - \frac {1} {7}
(x + 1)^7 = \\\frac {7} {56} (x + 1)^8 - \frac {8} {56}
(x + 1)^7 =\\ \frac {1} {56} (x + 1)^7 (7
(x + 1) - 8) = \frac {1} {56} (x + 1)^7 (7 x - 1)\]

Answer 19

@perl