$$2nCr(0$$≤r≤$$2n)$$ is greatest when r is equal to

Question

$$2nCr(0$$≤r≤$$2n)$$ is greatest when r is equal to

Infinity Learn · Accepted Answer

We have,  $$2nCr$$ $$2nCr$$−$$1=2n$$−$$r+1r$$            (1)For$$  $$2nCr$$ to be greatest $$2n$$−$$r+1r$$≥$$1$$⇒ $$2n$$−$$r+1$$≥r⇒$$2r$$≤$$2n+1$$⇒r≤$$n+12$$                                           $$(2)From$$ $$(1),  $$2nCr+1$$ $$2nCr=2n$$−(r+1)+1r+1=2n$$−$$rr+1$$.For  $$2nCr$$ to be greatest $$2n$$−$$rr+1$$≤$$1$$⇒ $$2n$$−r≤$$r+1$$⇒$$2r$$≥$$2n$$−$$1$$⇒r≥n−$$12$$                                          $$(3)From$$ $$(2) and (3), we get n−$$12$$≤r≤$$n+12$$⇒ r $$=$$ n $$(since$$ r is a positive $$integer)Hence$$,  $$2nCr$$ is greatest when r $$=$$ n.

$^{2 n} C_{r} (0 \leq r \leq 2 n)$ is greatest when r is equal to

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2nCr(0≤r≤2n) is greatest when r is equal to

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$^{2 n} C_{r} (0 \leq r \leq 2 n)$ is greatest when r is equal to