Question

Let $a=min\{x^2+2x+3,xϵR\}andb=\underset{\theta \to 0}{lim}\frac{1-cos\theta }{{\theta }^2}$. The value of ${\Sigma }_{r=0}^n{a}^r.b^{n-r}$ is

• A.
• B.
• C.
• D.

Answer 1

The correct option is C

$a=min\{x^2+2x+3,xϵR\}=min\left\{\right(x+1{)}^2+2,xϵR\}=2andb=\underset{\theta \to 0}{lim}\frac{1-cos\theta }{{\theta }^2}=\underset{\theta \to 0}{lim}\frac{(1-cos\theta )(1+cos\theta )}{{\theta }^2(1+cos\theta )}=\frac{1}{2}\therefore {\Sigma }_{r=0}^n{a}^r.b^{n-r}=b^n{\Sigma }_{r=0}^n{\left(\frac{a}{b}\right)}^r={\left(\frac{1}{2}\right)}^n{\Sigma }_{r=0}^n(4{)}^r=\frac{1}{2^n}(1+4+4^2+\dots \dots +4^n)=\frac{1}{2^n}.1.\left(\frac{4^{n+1}-1}{4-1}\right)=\frac{4^{n+1}-1}{{3.2}^n}$