Question

5+10+15+ to +5n = 5n(n+1)/2 for k+1​

Answer 1

Answer:

.

Step-by-step explanation:

The given statement is

5+10+15+...+5n=\frac{5n(n+1)}{2}5+10+15+...+5n=

2

5n(n+1)

For n = 1

\begin{lgathered}5=\frac{5\cdot1(1+1)}{2}\\\\5=5\cdot1\\\\5=5=>\text{ True}\end{lgathered}

5=

2

5⋅1(1+1)

5=5⋅1

5=5=> True

Hence, the given statement is true for n = 1

Let the given statement is true for n = k

So, we have

P(k)=5+10+15+...+5k-5+5k=\frac{5k(k+1)}{2}P(k)=5+10+15+...+5k−5+5k=

2

5k(k+1)

Now, we have to prove that the statement is true for n = k+1

Substitute n = k+1

\begin{lgathered}5+10+15+...+\frac{5(k+1)(k+1+1)}{2}\\\\5+10+15+...+\frac{(5k+5)(k+2)}{2}\\\\5+10+15+...+\frac{5k^2+15k+10}{2}\\\\5+10+15+...+\frac{5k(k+3)}{2}+5\\\\5+10+15+...+\frac{5k(k+1}{2}+\frac{10k}{2}+5\\\\5+10+15+...+\frac{5k(k+1}{2}+5k+5\\\\P(k)+5(k+1)\end{lgathered}

5+10+15+...+

2

5(k+1)(k+1+1)

5+10+15+...+

2

(5k+5)(k+2)

5+10+15+...+

2

5k

2

+15k+10

5+10+15+...+

2

5k(k+3)

+5

5+10+15+...+

2

5k(k+1

+

2

10k

+5

5+10+15+...+

2

5k(k+1

+5k+5

P(k)+5(k+1)

It means that it is true for n = k+1 as well

Hence, from mathematical induction, we can say that

5+10+15+...+5n=\frac{5n(n+1)}{2}5+10+15+...+5n=

2

5n(n+1)

Answer 2

Answer:

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