Let ` a _(j) = (7)/(4) ((2)/(3)) ^( j -1), j in N`. lf `b _(j) = a _(j) ^(2) + a _(j),` sum of the infinite sereies formed by `b _(j) 's` is `(10 + alpha )` where `[ (1)/( alpha ) ]` is equal to ([] represent greatest integer function)

Answer

To solve the problem, we need to follow these steps:

1. **Define the sequence \( a_j \)**:
\[ a_j = \frac{7}{4} \left( \frac{2}{3} \right)^{j - 1}, \quad j \in \mathbb{N} \]

2. **Define the sequence \( b_j \)**:
\[ b_j = a_j^2 + a_j \]

3. **Calculate \( a_j^2 \)**:
\[ a_j^2 = \left( \frac{7}{4} \left( \frac{2}{3} \right)^{j - 1} \right)^2 = \frac{49}{16} \left( \frac{2}{3} \right)^{2(j - 1)} = \frac{49}{16} \left( \frac{2}{3} \right)^{2j - 2} \]

4. **Combine \( b_j \)**:
\[ b_j = a_j^2 + a_j = \frac{49}{16} \left( \frac{2}{3} \right)^{2j - 2} + \frac{7}{4} \left( \frac{2}{3} \right)^{j - 1} \]

5. **Factor out common terms**:
\[ b_j = \left( \frac{2}{3} \right)^{j - 1} \left( \frac{49}{16} \left( \frac{2}{3} \right)^{j - 1} + \frac{7}{4} \right) \]

6. **Rewrite \( b_j \)**:
\[ b_j = \left( \frac{2}{3} \right)^{j - 1} \left( \frac{49}{16} \left( \frac{2}{3} \right)^{j - 1} + \frac{28}{16} \right) = \left( \frac{2}{3} \right)^{j - 1} \left( \frac{49}{16} \left( \frac{2}{3} \right)^{j - 1} + \frac{28}{16} \right) \]

7. **Sum the infinite series \( \sum_{j=1}^{\infty} b_j \)**:
\[ S = \sum_{j=1}^{\infty} b_j = \sum_{j=1}^{\infty} \left( \frac{2}{3} \right)^{j - 1} \left( \frac{49}{16} \left( \frac{2}{3} \right)^{j - 1} + \frac{28}{16} \right) \]

8. **Separate the sums**:
\[ S = \sum_{j=1}^{\infty} \left( \frac{49}{16} \left( \frac{2}{3} \right)^{2(j - 1)} \right) + \sum_{j=1}^{\infty} \left( \frac{28}{16} \left( \frac{2}{3} \right)^{j - 1} \right) \]

9. **Use the formula for the sum of a geometric series**:
- For the first sum:
\[ \sum_{j=1}^{\infty} \left( \frac{49}{16} \left( \frac{2}{3} \right)^{2(j - 1)} \right) = \frac{\frac{49}{16}}{1 - \left( \frac{2}{3} \right)^2} = \frac{\frac{49}{16}}{1 - \frac{4}{9}} = \frac{\frac{49}{16}}{\frac{5}{9}} = \frac{49 \cdot 9}{16 \cdot 5} = \frac{441}{80} \]

- For the second sum:
\[ \sum_{j=1}^{\infty} \left( \frac{28}{16} \left( \frac{2}{3} \right)^{j - 1} \right) = \frac{\frac{28}{16}}{1 - \frac{2}{3}} = \frac{\frac{28}{16}}{\frac{1}{3}} = \frac{28 \cdot 3}{16} = \frac{84}{16} = \frac{21}{4} \]

10. **Combine the results**:
\[ S = \frac{441}{80} + \frac{21}{4} = \frac{441}{80} + \frac{420}{80} = \frac{861}{80} \]

11. **Set the sum equal to \( 10 + \alpha \)**:
\[ 10 + \alpha = \frac{861}{80} \implies \alpha = \frac{861}{80} - 10 = \frac{861 - 800}{80} = \frac{61}{80} \]

12. **Find \( \frac{1}{\alpha} \)**:
\[ \frac{1}{\alpha} = \frac{80}{61} \]

13. **Calculate the greatest integer function**:
\[ \lfloor \frac{80}{61} \rfloor = 1 \]

### Final Answer:
The answer is \( \lfloor \frac{1}{\alpha} \rfloor = 1 \).