More and more threes

Algebra Level 4

$3,33,333,\ldots, \underbrace{33333\ldots3}_{n \text{ number of 3's}}$

Given that the sum of the $n$ number of terms above is equal to

$\frac 1a (b^{n+1} - cn - d )$

for positive integers $a,b,c$ and $d$ . Find the value of $a+b+c+d+1$ .

The answer is 57.

1 solution

Prakhar Gupta
May 7, 2015

Here is a tricky part in this problem. Let's investigate that. $3+33+333 + \ldots$ We will try to convert it in the form so that we may create a $GP$ in it. $\dfrac{1}{3}(9+99+999+ \ldots)$ $\dfrac{1}{3}(10-1+10^{2}-1+10^{3} - 1 + \ldots)$ Now we have created a $GP$ . Now it can be summed easily using formula for finite $GP$ summation. $\dfrac{1}{3}(\sum_{r=1}^{n}( 10^{r} ) - n)$ $\dfrac{1}{3} (\dfrac{10(10^{n}-1}{10-1}) - n)$ Now we just have to manipulate the expression to get the expression in desired format. $\dfrac{1}{3}\dfrac{10^{n+1} - 9n-10}{9}$ $\dfrac{1}{27}(10^{n+1} - 9n-10)$

Perfectly done.Even I did it in the same way and got it exactly right.

Rama Devi - 6 years ago

me too the same way

Sriram Venkatesan - 5 years, 11 months ago

If , n =3 , a = 1 , b = 5 , c =85 , d =1 , e = 1 , Then my answer is 93. But why ? Please help me.

3 + 33 + 333 = 369

a/e = 1

( b )^(n+1) = 5 ^ 4 = 625

c * n = 255

d = 1

So , 1 * ( 625 - 255 - 1 ) = 369

Shohag Hossen - 5 years, 11 months ago

Good solution.It is upvoted.

Sai Ram - 5 years, 11 months ago

More and more threes

The answer is 57.

1 solution

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