if $$a1$$,$$a2$$,$$a3a$$,$$4$$ are the coefficients of any four consecutive terms in the expansion of $$(1+x)n$$, then $$a1a1+a2+a3a3+a4$$ is equal to

Question

if $$a1$$,$$a2$$,$$a3a$$,$$4$$ are the coefficients of any four consecutive terms in the expansion of $$(1+x)n$$, then $$a1a1+a2+a3a3+a4$$ is equal to

Infinity Learn · Accepted Answer

Let $$a1$$,$$a2$$, $$a3$$, $$a4$$ be respectively the coefficients $$of(r$$ $$+$$ $$1)th$$, $$(r$$ $$+$$ $$21)th$$, $$(r$$ $$+$$ $$3)th$$ and $$(r$$ $$+$$ $$4)th$$ terms in theexpansion of $$(1$$ $$+$$ ,$$x)n$$ .Then,$$a1=nCr$$,   $$a2=nCr+1$$,  $$a3=nCr+2$$,  $$a4=$$′$$Cr+3Now$$ $$a1a1+a2+a3a3+a4=$$ nCr $$nCr+nCr+1+$$ $$nCr+2$$ $$nCr+2+nCr+3$$                                     $$=$$ nCr $$n+1Cr+1+$$ $$nCr+2n+1Cr+3$$                         ∵$$nCr+nCr+1=n+1Cr+1=$$ $$nCrn+1r+1nCr+$$ $$nCr+2n+1r+3nCr+2$$            ∵$$nCr=nrn$$−$$1Cr$$−$$1=r+1n+1+r+3n+1=2(r+2)n+1=2$$ $$nCr+1$$ $$n+1Cr+2=2$$ $$nCr+1$$ $$nCr+1+nCr+2=2a2a2+a3$$

if a₁,a₂,a₃a,₄ are the coefficients of any four consecutive terms in the expansion of $(1 + x)^{n}$ , then $\frac{a_{1}}{a_{1} + a_{2}} + \frac{a_{3}}{a_{3} + a_{4}}$ is equal to

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if a1,a2,a3a,4 are the coefficients of any four consecutive terms in the expansion of (1+x)n, then a1a1+a2+a3a3+a4 is equal to

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if a₁,a₂,a₃a,₄ are the coefficients of any four consecutive terms in the expansion of $(1 + x)^{n}$ , then $\frac{a_{1}}{a_{1} + a_{2}} + \frac{a_{3}}{a_{3} + a_{4}}$ is equal to