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Determine the following sum $$\frac{5}{3 \cdot 6 }\cdot \frac{1}{4^2} + \frac{5\cdot 8}{3 \cdot 6 \cdot 9 }\cdot \frac{1}{4^3} + \frac{5 \cdot 8 \cdot 11}{3 \cdot 6 \cdot 9 \cdot 12}\cdot \frac{1}{4^4} + \frac{5\cdot8\cdot 11\cdot 14}{3 \cdot 6\cdot 9\cdot 12 \cdot 15}\cdot \frac{1}{4^5} + \dots \dots$$

I tried to use Generalised Binomial Theorem, but I am unable to find $x,y,r$

The Generalized Binomial Theorem $$(x+y)^{r}=\sum _{k=0}^{\infty }{r \choose k}x^{r-k}y^{k}$$ where the ${r \choose k }$ denotes the falling factorial. I notice that the factors in the numerator is increasing, hence I am unable to use it.

GraduateStudent
  • 1,332

2 Answers2

2

Hint: $$\frac{5 \cdot 8 \cdot 11 \cdot 14}{3 \cdot 6 \cdot 9 \cdot 12 \cdot 15}=\frac{1}{3 \cdot 5!}(1+\frac{2}{3})(2+\frac{2}{3})(3+\frac{2}{3})(4+\frac{2}{3})=\frac{(-1)^4}{3 \cdot 5!}\prod_{k=0}^3{\left(-\frac{5}{3}-k\right)}$$.

Aphelli
  • 34,439
1

As the denominator has one extra multiplicand

and the number of terms in denominator $=$ the exponent of $\dfrac14$

$$2S=\dfrac{2\cdot5}{3\cdot6}\left(\dfrac14\right)^2+\dfrac{2\cdot5\cdot8}{3\cdot6\cdot9}\left(\dfrac14\right)^3+\cdots$$

Now like Calculating $1+\frac13+\frac{1\cdot3}{3\cdot6}+\frac{1\cdot3\cdot5}{3\cdot6\cdot9}+\frac{1\cdot3\cdot5\cdot7}{3\cdot6\cdot9\cdot12}+\dots? $

$$2S+1=1+\dfrac{\dfrac23\cdot\dfrac53}{2!}\left(\dfrac14\right)^2+\cdots$$

$$=1+\dfrac{\left(-\dfrac23\right)\cdot\left(-\dfrac23-1\right)} {2!}\left(-\dfrac14\right)^2+\cdots$$

$$=\left(1-\dfrac14\right)^{-2/3}$$ using Binomial series

We can use Calculating $1+\frac13+\frac{1\cdot3}{3\cdot6}+\frac{1\cdot3\cdot5}{3\cdot6\cdot9}+\frac{1\cdot3\cdot5\cdot7}{3\cdot6\cdot9\cdot12}+\dots? $

to find $x,n$ to be $-\dfrac14-\dfrac23$ respectively in $$2S+1=(1+x)^n$$

lab bhattacharjee
  • 274,582