lf some three consecutive coefficients in the binomial expansion of $$(x$$ $$+$$ $$1)n$$ in powers of x are in the ratio $$2$$ : $$15$$ : $$70$$, then the average of these three coefficients is

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Infinity Learn · Accepted Answer

Given binomial is (x+$$ $$1)n$$, whose general term, is $$Tr+1=nCrxrAccording$$ to the question, we have nCr−$$1$$:nCr:$$nCr+1=2$$:$$15$$:$$70now$$, nCr−$$1$$ $$nCr=215$$⇒n!$$(r$$−$$1)!(n$$−$$r+1)!n!r!(n$$−$$r)!$$=215$$⇒rn−$$r+1=215$$⇒ $$15r=2n$$−$$2r+2$$⇒ $$2n$$−$$17r+2=0----i$$⇒  similarly,   nCr $$nCr+1=1570$$⇒n!r!(n$$−$$r)!n!(r+1)!(n$$−r−$$1)!$$=314$$⇒ $$r+1n$$−$$r=314$$⇒$$14r+14=3n$$−$$3r$$⇒ $$3n$$−$$17r$$−$$14=0----iiOn$$ solving Eqs. (i) and (ii), we getn−$$16=0$$⇒$$n=16$$ and $$r=2Now$$, the average $$=$$ $$16C1+16C2+16C33$$                           $$=16+120+5603=6963=232$$

lf some three consecutive coefficients in the binomial expansion of (x + 1)ⁿ in powers of x are in the ratio 2 : 15 : 70, then the average of these three coefficients is

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lf some three consecutive coefficients in the binomial expansion of (x + 1)n in powers of x are in the ratio 2 : 15 : 70, then the average of these three coefficients is

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lf some three consecutive coefficients in the binomial expansion of (x + 1)ⁿ in powers of x are in the ratio 2 : 15 : 70, then the average of these three coefficients is