Recent content by gaganspidey

Thanks very much vela & HallsofIvy, finally I get the hang of this thing, how clumsy of me not to think of it.

^ That went over my head :redface: I mean that's the question I am asking, how do I divide them ?

Homework Statement This is the given Theorem in my book, everything seems fine except that I cannot figure how they expanded (xn - an) Homework Equations The Binomial Theorem The Attempt at a Solution According to me (xn - an) = {[(x+a)-a]n - an} and expanding it would yield...

^ Got it ! Thanks !

One more thing, they say when a body is released from another body moving upwards, u will be negative [Taking upward as (-ve) & downward as (+ve)] because first it reaches the highest point, i.e, it slightly moves up & then proceeds falling downward. So in this case, we'll take u=(-ve), g...

@Doc Al Oh I see ! Yes I get the same answer by taking upward as positive & downward as negative. Just that in this case we take u=-12m/s, s=65 & g=+9.8m/s2 Then why do they confuse us by using different sign conventions when sticking to a standard one does the job. Thanks for your help...

Homework Statement Example 3.13 : Example 3.1 Homework Equations v=u+at S=ut+1/2at2 v2-u2=2aS The Attempt at a Solution I know the solution is in front of me, that's ok but I have a confusion regarding negative & positive signs. In both the examples something is moving vertically upwards...

That means I am through with this problem though I need to increase my speed a bit else... :rolleyes:

This is what I've been able to work out : \sum_{r=0}^{n} r C^n_r x^r y^{n-r} <--------(given) = \frac{n!}{(r-1)!(n-1-(r-1)!} x^r y^{n-r} <---------(cancelling the r's & adding & subtracting 1 & regrouping them) = \frac{(nx)}{x} \frac{(n-1)!}{(r-1)!(n-1-(r-1)!} x^r y^{n-r}...

For n=4 :- \sum_{r=0}^4 r\cdot C_r^4 \cdot x^r y^{4-r} = 0 + 4 \cdot x y^3 + 12 \cdot x^2 y^2 + 12\cdot x^3 y + 4 \cdot x^4 = 4x( y^3 + 3 \cdot x y^2 + 3 \cdot x^2 y + x^3) = 4x(x+y)^3 = 4x <--------(substituting the value x+y=1) = nx <--------(when r=1 to n) {Is this step...

Do you mean : - \sum_{r=0}^{n} r/(nx) nCr = \sum_{r=0}^{n} r/(nx)n!/r(r-1)!(n-r)! = \sum_{r=0}^{n} 1/(nx)n!/(r-1)!(n-r)! <-----(cancelling the r's) = 0 + 1/(nx) n!/(n-1)! + 1/(nx) n!/(n-2)! +...+ 1/(nx) n!/(n-1)! <----(putting r=0,1,2...n) = 1/x + (n-1)/x + (n-1)(n-2)/2x + ... +1/x

This went over my head :redface:

Sorry, I still have to learn derivation/differentiation.

Hmm...got to give it a try.

I wonder how I could apply this formula because we need a geometric progression for it,ie, a2/a1=a3/a2 whereas I am not getting LHS = RHS.