Σn=0∞xn2n+4nΣ_{n=0}^∞\frac{x^n}{2^n+4^n}Σn=0∞2n+4nxn എന്ന അനുക്രമത്തിന്ടെ അഭിസരണ അർദ്ധ വ്യാസം ? A2B3C4D6Answer: C. 4 Read Explanation: an=12n+4na_n=\frac{1}{2^n+4^n}an=2n+4n1an+1=12n+1+4n+1a_{n+1}=\frac{1}{2^{n+1}+4^{n+1}}an+1=2n+1+4n+111R=limn→∞∣12n+1+4n+1×2n+4n1∣\frac{1}{R}=\lim_{n \to ∞}|\frac{1}{2^{n+1}+4^{n+1}} \times \frac{2^n+4^n}{1}|R1=limn→∞∣2n+1+4n+11×12n+4n∣=limn→∞∣4n[1+12n]4n+1[1+12n+1]∣=\lim _{n \to ∞} |\frac{4^n[1+ \frac{1}{2^n}]}{4^{n+1}[1+\frac{1}{2^{n+1}}]}|=limn→∞∣4n+1[1+2n+11]4n[1+2n1]∣1R=14\frac{1}{R}=\frac{1}{4}R1=41R=4R=4R=4 Read more in App
