Sum of the series

Probability Level 3

( 1 + 1 3 + 1 ⋅ 3 3 ⋅ 6 + 1 ⋅ 3 ⋅ 5 3 ⋅ 6 ⋅ 9 + ⋯ ) 2 = ? \left( 1 + \frac{1}{3}+\frac{1 \cdot 3}{3 \cdot 6}+\frac{1 \cdot 3 \cdot 5}{3 \cdot 6 \cdot 9}+ \cdots \right)^2= \, ?


The answer is 3.

Sandeep Bhardwaj by Sandeep Bhardwaj , 28, India

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1 solution

Akshat Sharda
Dec 18, 2015

We know,

( 1 + x ) n = 1 + n x + n ( n − 1 ) 2 ! x 2 + n ( n − 1 ) ( n − 2 ) 3 ! x 3 + … (1+x)^n=1+nx+\frac{n(n-1)} {2!}x^2+\frac{n(n-1)(n-2)} {3!}x^3+\ldots

Now,

n x = 1 3 nx=\frac{1}{3} and n ( n − 1 ) 2 ! x 2 = 1 6 \frac{n(n-1)}{2!}x^2=\frac{1} {6}

By solving the above equations, we'll get x = − 2 3 x=-\frac{2}{3} and n = − 1 2 n=-\frac{1}{2} .

Therefore, ( 1 + x ) n = ( 1 3 ) − 1 2 = 3 1 2 (1+x)^n=\left(\frac{1} {3}\right)^{-\frac{1} {2}}=3^{\frac{1}{2}} , so our answer is ( 3 1 2 ) 2 = 3 \left(3^{\frac{1}{2}}\right)^2 = 3

4 Helpful 0 Interesting 0 Brilliant 0 Confused

How did you get the the idea the the series would exist by making first two terms equal to some value ?

Vishal Yadav - 4 years, 3 months ago

We can see that the general term of expression is 1 2 × 1 3 n × ( 2 n n ) \frac {1}{2} \times \frac{1}{3^n} \times \binom{2n}{n} from second term 1 3 \frac{1}{3} . Any further suggestions ?

Vishal Yadav - 4 years, 2 months ago

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n = 1 gives second term.

Vishal Yadav - 4 years, 2 months ago

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