Question

Can you use the principle of mathematical induction to find a formula for the sum of the first n terms of a sequence? b) Can you use the principle of mathematical induction to determine whether a given formula for the sum of the first n terms of a sequence is correct? c) Find a formula for the sum of the first n even positive integers, and prove it using mathematical induction.

Short answer

a)The sum of the first n terms of a sequence is ,\({S_n} = \frac{{n\left( {n + 1} \right)}}{2}\).

b)Thus, the sum of the even positive integers is ,\({S_n} = \frac{{n\left( {n + 1} \right)}}{2}\).

c) Thus, the sum of the even positive integers is , \({S_n} = n\left( {n + 1} \right)\).

Step-by-step solution

Mathematical Induction.

We know the statement of mathematical induction.

If the statement is true for

\(p\left( 1 \right)\)is true.

if we assume\(n = k\)is true then\(p\left( k \right)\)is true.

if we assume\(n = k + 1\)is true then all the value of n is true in the statement.

(a) use the principle of mathematical induction to find a formula for the sum of the first n terms of a sequence?

1) Solution.

Sum of the first term \( = 1 = \frac{{1\left( {1 + 1} \right)}}{2}\)

The Sum of the first 2 terms is,

[a1] \(\begin{aligned}{c} = 1 + 2 = 3\\ = \frac{{2\left( {2 + 1} \right)}}{2}\end{aligned}\)

Sum of first 3 term \( = 1 + 2 + 3 = 6\)

\( = \frac{{3\left( {3 + 1} \right)}}{2}\).

In general, we can write, the sum of the first n -terms.

\(\begin{aligned}{c} = 1 + 2 + 3 + .......... + n\\ = \frac{{n\left( {n + 1} \right)}}{2}\end{aligned}\)

i.e formula for the sum of first n terms if a sequence is,

\({S_n} = \frac{{n\left( {n + 1} \right)}}{2}\).

[a1]

(b) Can you use the principle of mathematical induction to determine whether a given formula for the sum of the first n terms of a sequence is correct?

1) Explanation.

In the given question we need to solve the mathematical induction method to solve our question.

2) Solution.

We prove it by using the mathematical induction method,

\({S_n} = \frac{{n\left( {n + 1} \right)}}{2}\)

For \(n = 1\)

\({S_1} = \frac{{1\left( {1 + 1} \right)}}{2} = 1\)

i.e for \(n = 1\)above formula is true.

for \(n = 2\)

\({S_2} = \frac{{2\left( {2 + 1} \right)}}{2} = 3\)

\( = 1 + 2\)

i.e for \(n = 2\) above formula is also true.

Let the above formula is true\(k - 1\).

i.e\({S_{k - 1}} = \frac{{k\left( {k - 1} \right)}}{2}\).

Now,

\(\begin{aligned}{c}{S_k} = {S_{k - 1}} + k\\ = \frac{{k\left( {k - 1} \right)}}{2} + k\\ = \frac{{{k^2} - k + 2k}}{2}\\ = \frac{{{k^2} + k}}{2} = \frac{{k\left( {k + 1} \right)}}{2}\end{aligned}\)

Thus we get,

If this formula is true for\(n = k - 1\).

Then it is also true for\(n = k\).

Hence mathematical induction, \({S_n} = \frac{{n\left( {n + 1} \right)}}{2}\) is true for every\(n \in N\).

\(\therefore \)it is true for first n terms also

i.e, \({S_n} = \frac{{n\left( {n + 1} \right)}}{2}\).

Hence, here we have to prove.

Step 4:-(c) Solve the sum of the first n even positive integers, and prove it using mathematical induction.

1) Explanation.

In the given question the sum of the first n even positive integers.

Now, here we prove it.

2) Concept.

The formula for the sum of first n even positive terms.

\(2 + 4 + 6 + - - - - + 2n\)( given series in A.P)

Here\(d = 4 - 2 = 6 - 4 = 2\)

\(a = 2\)

\({S_n} = \frac{n}{2}\left( {2.2 + 2\left( {n - 1} \right)} \right)\)

( sum n terms of in A.P\({S_n} = \frac{n}{2}\left( {2a + \left( {n - 1} \right)d} \right)\))

\(\begin{aligned}{c}{S_n} = \frac{n}{2}\left( {4 + 2\left( {n - 1} \right)} \right)\\ = \frac{n}{2}\left( {2n + n} \right) = n\left( {n + 1} \right)\end{aligned}\)

Formula from first n even positive terms is\({S_n} = n\left( {n + 1} \right)\).

3) Solution.

Now here we prove it by mathematical induction.

Here we solve our question,

For \(n = 1\)

\({S_1} = 1\left( {1 + 1} \right) = 2\)

i.e above formula is true for \(n = 2\) also.

let it is true for\(n = k\),

then we prove it is true\(n = k + 1\).

Hence by mathematical induction, it is true for all\(n \in N\).

Now, let \({S_n}\) is true\(n = k\).

i.e

\(\begin{aligned}{c}{S_k} = k\left( {k + 1} \right)\\ = 2 + 4 + 6 + - - - - + 2k\\{S_{k + 1}} = {S_k} + 2\left( {k + 1} \right)\\ = k\left( {k + 1} \right) + 2\left( {k + 1} \right)\\ = \left( {k + 1} \right)\left( {k + 2} \right)\\ = \left( {k + 1} \right)\left( {k + 1 + 1} \right)\end{aligned}\)

i.e it is true for\(n = k + 1\).

Hence by mathematical induction, it is true for all\(n \in N\).

Hence it is true for first n even positive terms.

i.e \({S_n} = n\left( {n + 1} \right)\) is the sum of first ‘n’ even positive terms