If for some positive integer n, the coefficients of three consecutive terms in the binomial expansion of $$(1$$ $$+$$ $$x)n$$ $$+$$ $$5$$ are in the ratio $$5$$ : $$10$$ : $$14$$, then theIargest coefficient in this expansion is

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Infinity Learn · Accepted Answer

Let the three consecutive terms in the binomial expansion of (1+x)n+5$$ are $$n+5Cr$$−$$1$$,$$n+5Cr$$ and $$n+5Cr+1$$ Now, according to the given information  $$n+5Cr$$−$$1$$:$$n+5Cr$$:$$n+5Cr+1=5$$:$$10$$:$$14$$⇒ $$(n+5)!(r$$−$$1)!(n$$−$$r+6)!:(n+5)!r!(n$$−$$r+5)!:(n+5)!(r+1)!(n$$−$$r+4)!$$=5$$:$$10$$:$$14$$   ⇒$$1(n$$−$$r+6)(n$$−$$r+5)$$:$$1r(n$$−$$r+5)$$:$$1(r+1)r=5$$:$$10$$:$$14$$  So,  rn−$$r+6=510$$ ⇒ $$2r=n$$−$$r+6$$⇒$$n+6=3r$$  and  $$r+1n$$−$$r+5=57$$⇒ $$7r+7=5n$$−$$5r+25$$⇒ $$5n+18=12rFrom$$ Eqs. (i) and (ii), we have n $$=6$$.So, the largest coefficient in the expansion is same as the greatest binomial coefficient $$=11C5$$ or $$11C6=11$$!$$5$$!$$6$$!$$=11$$×$$10$$×$$9$$×$$8$$×$$75$$×$$4$$×$$3$$×$$2=462$$

If for some positive integer n, the coefficients of three consecutive terms in the binomial expansion of (1 + x)n + 5 are in the ratio 5 : 10 : 14, then theIargest coefficient in this expansion is

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