Question

mathematics induction
can anyone solve this questions​

Answer 1

GIVEN :–

$\\ \implies\bf 5 + 15 + 45 + ......... + 5 {(3)}^{n - 1} = \dfrac{5}{2}({3}^{n} - 1)$

TO PROVE :–

• Prove this by Using mathematics induction.

SOLUTION :–

Step (1) :–

• Put n = 1 –

$\\ \implies\bf 5 {(3)}^{1 - 1} = \dfrac{5}{2}({3}^{1} - 1)$

$\\ \implies\bf 5 {(3)}^{0} = \dfrac{5}{2}({3} - 1)$

$\\ \implies\bf 5= \dfrac{5}{ \cancel2}( \cancel2)$

$\\ \implies\bf 5=5$

• Hence , It's true for n = 1 .

Step (2) :–

• Now Let's assume that it's true for n = k :–

$\\ \implies\bf 5 + 15 + 45 + ......... + 5 {(3)}^{k-1} = \dfrac{5}{2}({3}^{k} - 1) \: \: \: \: \: - - - - eq.(1)$

Step (3) :–

• Now we should have to prove for n = k+1 :–

$\\ \implies\bf 5 + 15 + 45 + ......... + 5 {(3)}^{k} = \dfrac{5}{2}({3}^{k + 1} - 1)$

• Now let's take L.H.S. –

$\\ \bf \:\: = \:\:{ \underbrace{\bf 5 + 15 + 45 + ......... + 5 {(3)}^{k - 1}}} + 5 {(3)}^{k}$

• Using eq.(1) –

$\\ \: \: = \:\: \bf \dfrac{5}{2}({3}^{k} - 1) + 5 {(3)}^{k}$

$\\ \: \: = \:\: \bf \dfrac{5.{3}^{k} }{2} - \dfrac{5}{2} + 5 {(3)}^{k}$

$\\ \: \: = \:\: \bf {3}^{k} \bigg(\dfrac{5}{2} + 5\bigg)- \dfrac{5}{2}$

$\\ \: \: = \:\: \bf {3}^{k} \bigg(\dfrac{5 + 10}{2} \bigg)- \dfrac{5}{2}$

$\\ \: \: = \:\: \bf {3}^{k} \bigg(\dfrac{15}{2} \bigg)- \dfrac{5}{2}$

$\\ \: \: = \:\: \bf {3}^{k} \bigg(\dfrac{5.3}{2} \bigg)- \dfrac{5}{2}$

$\\ \: \: = \:\: \bf {3}^{k + 1} \bigg(\dfrac{5}{2} \bigg)- \dfrac{5}{2}$

$\\ \: \: = \:\: \bf \dfrac{5}{2}({3}^{k + 1}-1)$

$\\ \: \: = \:\: \bf R.H.S.$

$\\ \implies \large{ \boxed{ \boxed{\bf Hence \: \: proved}}}$