Question

Exercițiul 2, VA ROG, DAU COROANA!
Măcar un subpunct, va rog!​

Answer 1

Răspuns:

Explicație pas cu pas:

a) 5¹¹⁺¹²⁺¹³⁺·······⁺³⁰ = (5ⁿ)³·25 <=>

5⁽³⁰⁺¹¹⁾⁽³⁰⁻¹¹⁺¹⁾÷²  = 5³ⁿ·5² <=>

5⁴¹ˣ¹⁰ = 5³ⁿ⁺² =>

3n+2 = 410 => 3n = 408 =>

n = 408:3 => n = 136

b) 7²⁰⁺²⁴⁺²⁸⁺····⁺⁴⁸ = 49ⁿ <=>

7⁽²⁰⁺⁴⁸⁾⁽⁽⁴⁸⁻²⁰⁾÷⁴⁺¹⁾÷² = 7²ⁿ =>

68·8:2 = 2n => 2n = 272 =>

n = 272:2 => n = 136

c) 13¹⁺³⁺⁵⁺⁷⁺···⁺⁴⁹ = 13ⁿ^² =>

1+3+5+7+...+49 = n² =>

(1+49)·[(49-1):2+1]:2 = n² =>

n² = 25·25 => n = 25

d) 5ⁿ⁺²+5ⁿ⁺¹ - 15ⁿ:3ⁿ = 29·25¹³ <=>

5²·5ⁿ+5·5ⁿ - (15:3)ⁿ = 29·(5²)¹³ =>

25·5ⁿ+5·5ⁿ - 5ⁿ = 29·5²⁶ =>

5ⁿ(25+5-1) =  29·5²⁶ =>

29·5ⁿ =  29·5²⁶ =>

n = 26

e) 9ⁿ⁺¹·9²ⁿ⁺² = 27¹⁴ =>

9ⁿ⁺¹⁺²ⁿ⁺² = (3³)¹⁴ =>

9³ⁿ⁺³ = 3³ˣ¹⁴ =>

3²⁽³ⁿ⁺³⁾ = 3³ˣ¹⁴ =>

2·3·(n+1) = 3·14  I:6  =>

n+1 = 7 => n = 6

Answer 2

Explicație pas cu pas:

a)

${5}^{11 + 12 + 13 + ... + 30} = {({5}^{n})}^{3} \cdot 25 \\ {5}^{( \frac{30 \times 31}{2} - \frac{10 \times 11}{2} )} = {5}^{3n} \cdot {5}^{2} \\ {5}^{410} = {5}^{3n + 2} \iff 3n + 2 = 410 \\ 3n = 408 \implies n = 136$

b)

${7}^{20 + 24 + 28 + ... + 48} = {49}^{n} \\ {7}^{4 \cdot (5 + 6 + 7 + ... + 12)} = {( {7}^{2} )}^{n} \\ {7}^{4 \cdot ( \frac{12 \cdot 13}{2} - \frac{4 \cdot 5}{2})} = {7}^{2n} \\ 2n = 4 \cdot 68 \implies n = 136$

c)

${13}^{1 + 3 + 5 + 7 + ... + 49} = {13}^{ {n}^{2} } \\ 1 + 3 + 5 + 7 + ... + 49 = {n}^{2} \\ {25}^{2} = {n}^{2} \implies n = 25$

d)

${5}^{n + 2} + {5}^{n + 1} - {15}^{n} : {3}^{n} = 29 \cdot {25}^{13} \\ {5}^{n + 2} + {5}^{n + 1} - {5}^{n} = 29 \cdot {( {5}^{2} )}^{13} \\ {5}^{n} \cdot ({5}^{2} + 5 - 1) = 29 \cdot {5}^{26} \\ {5}^{n} \cdot 29 = 29 \cdot {5}^{26} \\ {5}^{n} = {5}^{26} \implies n = 26$

e)

${9}^{n + 1} \cdot {9}^{2n + 2} = {27}^{14} \\ {( {3}^{2} )}^{n + 1} \cdot {( {3}^{2} )}^{2n + 2} = {( {3}^{3} )}^{14} \\ {3}^{2n + 2} \cdot {3}^{4n + 4} = {3}^{42} \\ {3}^{2n + 2 + 4n + 4} = {3}^{42} \iff 6n + 6 = 42 \\ 6n = 36 \implies n = 6$