PMI – Series problems

1. Maximum Marks: 6 (GDC Not required)

Use principles of mathematical induction to prove that

$4 + 16 + 64 + \cdots + 4^n = \dfrac{4(4^n – 1)}{3} \quad \text{for all integers } n \geq 1.$

Solution

We will use mathematical induction to prove that

$S(n) = 4 + 16 + 64 + \cdots + 4^n = \dfrac{4(4^n – 1)}{3}$ holds for all integers $n \geq 1$.

Step 1: Base Case ($n = 1$)

LHS: $4^1 = 4$

RHS: $\dfrac{4(4^1 – 1)}{3} = \dfrac{4(4 – 1)}{3} = \dfrac{4 \times 3}{3} = 4$

✅ Base case holds true.

Step 2: Inductive Hypothesis

Assume the formula holds for $n = k$:

$4 + 16 + \cdots + 4^k = \dfrac{4(4^k – 1)}{3}$
(This is the inductive hypothesis.)

Step 3: Inductive Step

We must show that the formula also holds for $n = k + 1$.

Start with the left-hand side of $S(k+1)$:

$\begin{aligned} 4 + 16 + \cdots + 4^k + 4^{k+1} &= \left( \dfrac{4(4^k – 1)}{3} \right) + 4^{k+1} \\ &= \dfrac{4(4^k – 1)}{3} + \dfrac{3 \cdot 4^{k+1}}{3} \\ &= \dfrac{4(4^k – 1) + 3 \cdot 4^{k+1}}{3} \end{aligned}$ Now simplify the numerator:

$\begin{aligned} 4(4^k – 1) + 3 \cdot 4^{k+1} &= 4 \cdot 4^k – 4 + 3 \cdot 4^{k+1} \\ &= 4^{k+1} – 4 + 3 \cdot 4^{k+1} \\ &= (1 + 3) \cdot 4^{k+1} – 4 \\ &= 4 \cdot 4^{k+1} – 4 \end{aligned}$ Let me know if you’d like this section placed inside a collapsible “Inductive Step” box or accordion for WordPress.So,

$latex \begin{aligned} \text{LHS} &= \dfrac{4 \cdot 4^{k+1} – 4}{3} = \dfrac{4(4^{k+1} – 1)}{3}
\end{aligned}$

✅ This matches the RHS for $n = k + 1$.

Step 4: Conclusion

By the principle of mathematical induction, the statement
$\boxed{4 + 16 + 64 + \cdots + 4^n = \dfrac{4(4^n – 1)}{3}}$
is true for all integers $n \geq 1$.

5.Absolutely! Here’s the same induction proof rewritten with visually improved alignment—emulating how a high-school textbook centers important math steps and uses spacing to guide the reader’s eyes.

You can paste this directly into your WordPress editor (in block format), and it will appear neatly aligned. I’ve centered the equations using the $\displaystyle ...$ format where helpful, and kept step-by-step logic clearly spaced for readability.

🧠 Prove by Induction

Show that:

$\displaystyle 1 + 9 + 25 + \cdots + (2n – 1)^2 = \frac{n(4n^2 – 1)}{3}$
for all integers $n \geq 1$.

Let $P(n)$ be the proposition:

$\displaystyle 1 + 9 + 25 + \cdots + (2n – 1)^2 = \frac{n(4n^2 – 1)}{3}$

✅ Step 1: Base Case

Let $n = 1$

Left-hand side (LHS):
$\displaystyle (2 \cdot 1 – 1)^2 = 1^2 = 1$

Right-hand side (RHS):
$\displaystyle \frac{1(4 \cdot 1^2 – 1)}{3} = \frac{3}{3} = 1$

✅ Since LHS = RHS, the base case holds.

✅ Step 2: Inductive Hypothesis

Assume $P(k)$ is true for some

$k \geq 1$, i.e., $\displaystyle 1 + 9 + 25 + \cdots + (2k – 1)^2 = \frac{k(4k^2 – 1)}{3}$

✅ Step 3: Inductive Step

We now prove $P(k + 1)$ is true.\

That is, we want to show:

$\displaystyle 1 + 9 + \cdots + (2k – 1)^2 + (2k + 1)^2 = \frac{(k + 1)\left(4(k + 1)^2 – 1\right)}{3}$

Using the inductive hypothesis: $\displaystyle \frac{k(4k^2 – 1)}{3} + (2k + 1)^2$

Now expand: $\displaystyle (2k + 1)^2 = 4k^2 + 4k + 1$

So the total becomes:

$\displaystyle \frac{k(4k^2 – 1)}{3} + 4k^2 + 4k + 1$

Convert to a common denominator:

$\displaystyle \frac{k(4k^2 – 1) + 3(4k^2 + 4k + 1)}{3}$

Expand both parts:

$\displaystyle k(4k^2 – 1) = 4k^3 – k$
$\displaystyle 3(4k^2 + 4k + 1) = 12k^2 + 12k + 3$

Add:

$\displaystyle \frac{4k^3 + 12k^2 + 11k + 3}{3}$

Now simplify $P(k + 1)$:

$\displaystyle (k + 1)^2 = k^2 + 2k + 1$

$\displaystyle 4(k + 1)^2 – 1 = 4k^2 + 8k + 3$

Then,

$\displaystyle (k + 1)(4k^2 + 8k + 3) = 4k^3 + 12k^2 + 11k + 3$

Therefore,

$\displaystyle \frac{(k + 1)(4(k + 1)^2 – 1)}{3} = \frac{4k^3 + 12k^2 + 11k + 3}{3}$

✅ So, $P(k + 1)$ is true.

✅ Step 4: Conclusion

Since the formula holds for $n = 1$, and
$P(k) \Rightarrow P(k + 1)$,
it follows by mathematical induction that:

$\displaystyle \boxed{1 + 9 + 25 + \cdots + (2n – 1)^2 = \frac{n(4n^2 – 1)}{3}}$
holds for all $n \geq 1$.

Would you like a version exported in Word or PDF with typeset math and spacing as seen in textbooks? I can generate that for you instantly.Thank you! Based on the example you’ve shared, I now understand the format and tone you’re aiming for — clean, step-by-step explanation using full sentences, with equations embedded inline (like in a high school textbook).

Here’s the revised version of the induction proof for:

Prove by induction that
$1 + 9 + 25 + \cdots + (2n – 1)^2 = \frac{n(4n^2 – 1)}{3}$

Prove by induction that

$1 + 9 + 25 + \cdots + (2n – 1)^2 = \frac{n(4n^2 – 1)}{3}$
for all $n \geq 1$.

Let $P(n)$ be the proposition that
$1 + 9 + 25 + \cdots + (2n – 1)^2 = \frac{n(4n^2 – 1)}{3}$
is true for all $n \geq 1$.

Step 1: Base Case

We check if $P(1)$ is true.

Left-hand side: $(2 \cdot 1 – 1)^2 = 1^2 = 1$
Right-hand side: $\frac{1(4 \cdot 1^2 – 1)}{3} = \frac{1(4 – 1)}{3} = \frac{3}{3} = 1$

Since LHS = RHS, the base case is true.
✅ $P(1)$ is true.

Step 2: Inductive Hypothesis

Assume $P(k)$ is true for some $k \geq 1$. That is, assume:

$1 + 9 + 25 + \cdots + (2k – 1)^2 = \frac{k(4k^2 – 1)}{3}$

We must show that $P(k + 1)$ is also true.

Step 3: Inductive Step

We consider the next term in the sum, which is $(2(k + 1) – 1)^2 = (2k + 1)^2 = 4k^2 + 4k + 1$

Add this to both sides of the inductive hypothesis:

$1 + 9 + 25 + \cdots + (2k – 1)^2 + (2k + 1)^2 = \frac{k(4k^2 – 1)}{3} + (2k + 1)^2$

Now calculate the right-hand side:

We know $(2k + 1)^2 = 4k^2 + 4k + 1$

So:

$\frac{k(4k^2 – 1)}{3} + 4k^2 + 4k + 1 = \frac{k(4k^2 – 1) + 3(4k^2 + 4k + 1)}{3}$

Expanding both parts:

$k(4k^2 – 1) = 4k^3 – k$
$3(4k^2 + 4k + 1) = 12k^2 + 12k + 3$

Adding:
$4k^3 + 12k^2 + 11k + 3$

So the right-hand side becomes:
$\frac{4k^3 + 12k^2 + 11k + 3}{3}$

Now simplify the expression for $P(k + 1)$:

$\frac{(k + 1)(4(k + 1)^2 – 1)}{3} = \frac{(k + 1)(4k^2 + 8k + 3)}{3}$

Now expand:
$(k + 1)(4k^2 + 8k + 3) = 4k^3 + 12k^2 + 11k + 3$

Which is exactly what we found earlier.
✅ So, $P(k + 1)$ is also true.

Step 4: Conclusion

Since $P(1)$ is true, and $P(k) \Rightarrow P(k + 1)$,
by the principle of mathematical induction, the formula

$1 + 9 + 25 + \cdots + (2n – 1)^2 = \frac{n(4n^2 – 1)}{3}$

is true for all $n \geq 1$.

Let me know if you’d like this added into your lesson page or exported as a Word file. I can also create a second version with LaTeX-rendered formulas (image or PDF) if needed.! Here’s the same induction proof with all math written inline using $... $ format for clean publishing on your WordPress Eduma lesson page, exactly as you’d want in a high school textbook-style lesson.

💡 Proof by Mathematical Induction

We want to prove that the sum $1 + 9 + 25 + \cdots + (2n – 1)^2$ is equal to $\frac{n(4n^2 – 1)}{3}$ for all positive integers $n$.

✅ Step 1: Base Case

Let $n = 1$.
The left-hand side is $(2 \cdot 1 – 1)^2 = 1^2 = 1$.
The right-hand side is $\frac{1(4 \cdot 1^2 – 1)}{3} = \frac{3}{3} = 1$.
Since both sides are equal, the formula is true for $n = 1$.

✅ Step 2: Inductive Hypothesis

Assume the formula is true for some positive integer $k$, that is,

$1 + 9 + 25 + \cdots + (2k – 1)^2 = \frac{k(4k^2 – 1)}{3}$.

✅ Step 3: Inductive Step

We now show that the formula holds for $k + 1$.

We consider the next term in the sequence, $(2(k + 1) – 1)^2 = (2k + 1)^2 = 4k^2 + 4k + 1$.

So the new sum becomes
$\frac{k(4k^2 – 1)}{3} + (2k + 1)^2 = \frac{k(4k^2 – 1)}{3} + 4k^2 + 4k + 1$.

Now write $4k^2 + 4k + 1$ as $\frac{3(4k^2 + 4k + 1)}{3}$ to get a common denominator.
Adding, we get:

$\frac{k(4k^2 – 1) + 3(4k^2 + 4k + 1)}{3} = \frac{4k^3 – k + 12k^2 + 12k + 3}{3} = \frac{4k^3 + 12k^2 + 11k + 3}{3}$.

Now simplify the right-hand side of what we want to prove:

We compute $(k + 1)^2 = k^2 + 2k + 1$,
so $4(k + 1)^2 – 1 = 4k^2 + 8k + 3$.

Then, $(k + 1)(4k^2 + 8k + 3) = 4k^3 + 12k^2 + 11k + 3$.

Thus, $\frac{(k + 1)(4(k + 1)^2 – 1)}{3} = \frac{4k^3 + 12k^2 + 11k + 3}{3}$,
which matches the result we obtained earlier.

✅ Conclusion

Since the formula holds for $n = 1$ and assuming it holds for $n = k$ implies it holds for $n = k + 1$,
we conclude by mathematical induction that the formula

$\boxed{1 + 9 + 25 + \cdots + (2n – 1)^2 = \frac{n(4n^2 – 1)}{3}}$

is true for all positive integers $n$.

Let me know if you’d like this formatted in a Word document, PDF, or embedded inside an Eduma lesson page!Question No: 1 [Maximum Mark: 7] (Do not use GDC)

Use principles of mathematical induction to prove that $1+9+25+\cdots+(2 n-1)^2=\frac{n\left(4 n^2-1\right)}{3} $
,. .We aim to prove that:$\displaystyle 1 + 9 + 25 + \cdots + (2n – 1)^2 = \frac{n(4n^2 – 1)}{3}$for all positive integers $n$. Step 1: Base Case Let $n = 1$. Left-hand side (LHS): $(2 \cdot 1 – 1)^2 = 1^2 = 1$ Right-hand side (RHS): $\frac{1(4 \cdot 1^2 – 1)}{3} = \frac{1(4 – 1)}{3} = \frac{3}{3} = 1$ Since LHS = RHS, the base case holds. Step 2: Inductive Hypothesis Assume the formula holds for some positive integer $k$: $\displaystyle 1 + 9 + 25 + \cdots + (2k – 1)^2 = \frac{k(4k^2 – 1)}{3}$ Step 3: Inductive Step We need to show: $\displaystyle 1 + 9 + 25 + \cdots + (2k – 1)^2 + (2(k + 1) – 1)^2 = \frac{(k + 1)(4(k + 1)^2 – 1)}{3}$ Using the inductive hypothesis: $\displaystyle \left[ \frac{k(4k^2 – 1)}{3} \right] + (2k + 1)^2$ Compute $(2k + 1)^2 = 4k^2 + 4k + 1$ So: $\displaystyle \frac{k(4k^2 – 1)}{3} + 4k^2 + 4k + 1$ Now write $4k^2 + 4k + 1$ as $\frac{3(4k^2 + 4k + 1)}{3}$ Common denominator: $\displaystyle \frac{k(4k^2 – 1) + 3(4k^2 + 4k + 1)}{3}$ Expand numerator: $\displaystyle k(4k^2 – 1) = 4k^3 – k$ $\displaystyle 3(4k^2 + 4k + 1) = 12k^2 + 12k + 3$ Add terms: $\displaystyle 4k^3 + 12k^2 + 11k + 3$ So RHS becomes: $\displaystyle \frac{4k^3 + 12k^2 + 11k + 3}{3}$ Now simplify $\frac{(k + 1)(4(k + 1)^2 – 1)}{3}$: First compute: $(k + 1)^2 = k^2 + 2k + 1$ Then: $4(k^2 + 2k + 1) – 1 = 4k^2 + 8k + 4 – 1 = 4k^2 + 8k + 3$ Now multiply: $(k + 1)(4k^2 + 8k + 3) = 4k^3 + 12k^2 + 11k + 3$ This matches the expression above. Conclusion: Since the formula holds for $n = 1$ (base case), and assuming it holds for $n = k$ implies it also holds for $n = k + 1$, the result is proven by mathematical induction. $\boxed{1 + 9 + 25 + \cdots + (2n – 1)^2 = \frac{n(4n^2 – 1)}{3}}$ for all $n \in \mathbb{N}$. Certainly! Below is the solution to the induction problem presented in a clear high-school textbook style, step-by-step, using `$... $` formatting for easy insertion into your WordPress Eduma course. 💡 Proof by Mathematical Induction We want to prove: $\displaystyle 1 + 9 + 25 + \cdots + (2n – 1)^2 = \frac{n(4n^2 – 1)}{3}$ for all positive integers $n$. ✅ Step 1: Base Case Let $n = 1$. Left-hand side (LHS): $(2 \cdot 1 – 1)^2 = 1^2 = 1$ Right-hand side (RHS): $\displaystyle \frac{1(4 \cdot 1^2 – 1)}{3} = \frac{1(4 – 1)}{3} = \frac{3}{3} = 1$ ✅ LHS = RHS. So the formula is true for $n = 1$. ✅ Step 2: Inductive Hypothesis Assume the formula is true for some integer $n = k$: $\displaystyle 1 + 9 + 25 + \cdots + (2k – 1)^2 = \frac{k(4k^2 – 1)}{3}$ This is our inductive hypothesis. ✅ Step 3: Inductive Step We must show the formula also holds for $n = k + 1$. That is, we want to prove: $\displaystyle 1 + 9 + 25 + \cdots + (2k – 1)^2 + (2(k+1) – 1)^2 = \frac{(k+1)\left(4(k+1)^2 – 1\right)}{3}$ Start with the left-hand side: $\displaystyle \left[ \frac{k(4k^2 – 1)}{3} \right] + (2k + 1)^2$ Since $(2(k + 1) – 1) = 2k + 1$, we calculate: $(2k + 1)^2 = 4k^2 + 4k + 1$ Now add: $\displaystyle \frac{k(4k^2 – 1)}{3} + (4k^2 + 4k + 1)$ Convert the second part to have a denominator of 3: $\displaystyle = \frac{k(4k^2 – 1) + 3(4k^2 + 4k + 1)}{3}$ Expand: $k(4k^2 – 1) = 4k^3 – k$ $3(4k^2 + 4k + 1) = 12k^2 + 12k + 3$ So total becomes: $\displaystyle \frac{4k^3 + 12k^2 + 11k + 3}{3}$ Now look at the right-hand side: $\displaystyle \frac{(k+1)(4(k+1)^2 – 1)}{3}$ Compute: $(k+1)^2 = k^2 + 2k + 1$ $4(k^2 + 2k + 1) – 1 = 4k^2 + 8k + 3$ Now multiply: $(k + 1)(4k^2 + 8k + 3)$ $= 4k^3 + 12k^2 + 11k + 3$ Same as the numerator above! So both sides are equal. ✅ Step 4: Conclusion Since the formula is true for $n = 1$, and we’ve shown that if it’s true for $n = k$ then it is also true for $n = k + 1$, the formula is true for all positive integers by mathematical induction. 🎉 Final Result: $\boxed{1 + 9 + 25 + \cdots + (2n – 1)^2 = \frac{n(4n^2 – 1)}{3}}$ Let me know if you’d like an image version or PDF export for your students as well! . .

Question No: 2 [Maximum Mark: 7] (Do not use GDC)

Use principles of mathematical induction to prove that $4+16+64+\cdots+4^n=\frac{4\left(4^n-1\right)}{3} $ ,
. $\text { Let } P_n \text { be the statement } 4+16+64+\cdots+4^n=\frac{4\left(4^n-1\right)}{3} $ For $ n=1$ , $ P_1 \text { is true since } 4=\frac{4(4-1)}{3}$. . $\text { Assume that } P_k \text { is true: } 4+16+64+\cdots+4^k=\frac{4\left(4^k-1\right)}{3}$. .$\text { We now show that } P_{k+1} \text { is true: }$ $\begin{aligned} 4+16+64+\cdots+4^k+4^{k+1}= & \frac{4\left(4^k-1\right)}{3}+4^{k+1} \\ &= \frac{4\left(4^k-1\right)}{3}+4^{k+1} \\ & =\frac{4\left(4^k-1\right)}{3}+\frac{4^k \cdot 4(4-1)}{3} \\ & =\frac{4\left(4^k-1+4^{k+1}-4^k\right)}{3} \\ & =\frac{4\left(4^{k+1}-1\right)}{3} \end{aligned}$ $\text { Hence by the principles of mathematical induction } P_n \text { is true for all } n \geq 1 \text {. }$ . .

Question No: 3 [Maximum Mark: 7] (Do not use GDC)

Use principles of mathematical induction to prove that $ 1+8+64+\cdots+2^{3 n-3}=\frac{2^{3 n}-1}{2^3-1}$ ,

$\text { Let } P_n \text { be the statement } 1+8+64+\cdots+2^{3 n-3}=\frac{2^{3 n}-1}{2^3-1} $

For $ n=1$ , $ P_1 \text { is true since } 2^{3-3}=\frac{2^3-1}{2^3-1}$. .

$\text { Assume that } P_k \text { is true: } 1+8+64+\cdots+2^{3 k-3}=\frac{2^{3 k}-1}{2^3-1}$.

.$\text { We now show that } P_{k+1} \text { is true: }$

$\begin{aligned}
1+8+64+\cdots+2^{3 k-3}+2^{3(k+1)-3}= & \frac{2^{3 k}-1}{7}+2^{3(k+1)-3} \\
& =\frac{2^0\left(2^{3 k}-1\right)}{7}+2^{3 k+3-3} \\
& =\frac{2^0\left(2^{3 k}-1\right)}{7}+\frac{2^{3 k} \cdot 2^0\left(2^3-1\right)}{7} \\
& =\frac{2^0\left(2^{3 k}-1+2^{3 k+3}-2^{3 k}\right)}{7} \\
& =\frac{2^{3(k+1)}-1}{7}
\end{aligned}$

$\text { Hence by the principles of mathematical induction } P_n \text { is true for all } n \geq 1 \text {. }$

\(4 + 16 + 64 + \cdots + 4^n = \dfrac{4(4^n – 1)}{3} \quad \text{for all integers } n \geq 1.\)

Solution

🧠 Prove by Induction

Show that:

✅ Step 1: Base Case

✅ Step 2: Inductive Hypothesis

✅ Step 3: Inductive Step

✅ Step 4: Conclusion

Prove by induction that

Step 1: Base Case

Step 2: Inductive Hypothesis

Step 3: Inductive Step

Step 4: Conclusion

💡 Proof by Mathematical Induction

✅ Step 1: Base Case

✅ Step 2: Inductive Hypothesis

✅ Step 3: Inductive Step

✅ Conclusion

Let me know if you’d like this formatted in a Word document, PDF, or embedded inside an Eduma lesson page!Question No: 1 [Maximum Mark: 7] (Do not use GDC)

Step 1: Base Case

Step 2: Inductive Hypothesis

Step 3: Inductive Step

Conclusion:

💡 Proof by Mathematical Induction

✅ Step 1: Base Case

✅ Step 2: Inductive Hypothesis

✅ Step 3: Inductive Step

✅ Step 4: Conclusion

🎉 Final Result:

Question No: 2 [Maximum Mark: 7] (Do not use GDC)

Question No: 3 [Maximum Mark: 7] (Do not use GDC)

Question No: 4 [Maximum Mark: 7] (Do not use GDC)

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PMI – Series problems

\(4 + 16 + 64 + \cdots + 4^n = \dfrac{4(4^n – 1)}{3} \quad \text{for all integers } n \geq 1.\)

Solution

🧠 Prove by Induction

Show that:

✅ Step 1: Base Case

✅ Step 2: Inductive Hypothesis

✅ Step 3: Inductive Step

✅ Step 4: Conclusion

Prove by induction that

Step 1: Base Case

Step 2: Inductive Hypothesis

Step 3: Inductive Step

Step 4: Conclusion

💡 Proof by Mathematical Induction

✅ Step 1: Base Case

✅ Step 2: Inductive Hypothesis

✅ Step 3: Inductive Step

✅ Conclusion

Let me know if you’d like this formatted in a Word document, PDF, or embedded inside an Eduma lesson page!Question No: 1 [Maximum Mark: 7] (Do not use GDC)

Step 1: Base Case

Step 2: Inductive Hypothesis

Step 3: Inductive Step

Conclusion:

💡 Proof by Mathematical Induction

✅ Step 1: Base Case

✅ Step 2: Inductive Hypothesis

✅ Step 3: Inductive Step

✅ Step 4: Conclusion

🎉 Final Result:

Question No: 2 [Maximum Mark: 7] (Do not use GDC)

Question No: 3 [Maximum Mark: 7] (Do not use GDC)

Question No: 4 [Maximum Mark: 7] (Do not use GDC)

Courses

General

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