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@ganeshie8 \[\huge \left(\begin{matrix}n+2 \\ 2\end{matrix}\right)\] does this evaluate to 2n+1
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\[\dfrac{(n-2)!}{(n-4)!2!}\]
\[\dfrac{(n-2)(n-3)(n-4)!}{(n-4)!2!}\]
(n-2)(n-3)/2!
yeah you get a square term
it cannot be simplified firther right?
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further*
it can never become 2n+1 http://www.wolframalpha.com/input/?i=%28n-2%29+choose+2
okay
hos did u get \(\binom{n-2}{2}\) ?
there is a formula
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|dw:1417932170736:dw|
k number of terms
for (a+b+c)^n that formula should give you \[\binom{n+3-1}{3-1}\]
which is same as \[\binom{n+2}{2}\]
right ?
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yes
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