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10765 Questions
1
Prove by induction that $1+4+7+\cdots+(3n-2)=\dfrac{n(3n-1)}{2}$ for $n\in\mathbb{Z}^+$
▾
✦ Full Worked Solution
1
Step 1 — Base Case (n = 1)
Verify the statement holds for $n=1$
\[\text{LHS}=1,\quad\text{RHS}=\dfrac{1(3-1)}{2}=1\quad\checkmark\]
2
Step 2 — Inductive Hypothesis
Assume true for $n=k$ :
\[1+4+\cdots+(3k-2)=\dfrac{k(3k-1)}{2}\]
3
Step 3 — Inductive Step
Add the next term $(3k+1)$ :
\[\dfrac{k(3k-1)}{2}+(3k+1)=\dfrac{3k^2-k+6k+2}{2}=\dfrac{3k^2+5k+2}{2}=\dfrac{(k+1)(3k+2)}{2}\quad\checkmark\]
4
Step 4 — Conclusion
By the Principle of Mathematical Induction, the statement is true for all $n\in\mathbb{Z}^+$
✅ Answer
Proved for all $n\in\mathbb{Z}^+$
2
Prove by mathematical induction that $1+2+4+\cdots+2^{n-1}=2^n-1$ for all integers $n\ge1$
▾
✦ Full Worked Solution
1
Step 1 — Base Case (n = 1)
\[\text{LHS}=1,\quad\text{RHS}=2-1=1\quad\checkmark\]
2
Step 2 — Inductive Hypothesis
Assume true for $n=k$ :
\[1+2+\cdots+2^{k-1}=2^k-1\]
3
Step 3 — Inductive Step
Add $2^k$ :
\[(2^k-1)+2^k=2\cdot2^k-1=2^{k+1}-1\quad\checkmark\]
4
Step 4 — Conclusion
By Mathematical Induction, true for all integers $n\ge1$
✅ Answer
Proved for all integers $n\ge1$
3
Prove by mathematical induction that $T_1+(T_1+d)+\cdots+[T_1+(n-1)d]=\dfrac{n}{2}[2T_1+(n-1)d]$ for all integers $n\ge1$
▾
✦ Full Worked Solution
1
Step 1 — Base Case (n = 1)
\[\text{LHS}=T_1,\quad\text{RHS}=\tfrac{1}{2}[ \, 2T_1 \, ]=T_1\quad\checkmark\]
2
Step 2 — Inductive Hypothesis
Assume true for $n=k$ :
\[T_1+(T_1+d)+\cdots+[ \, T_1+(k-1)d \, ]=\dfrac{k}{2}[ \, 2T_1+(k-1)d \, ]\]
3
Step 3 — Inductive Step
Add $[ \, T_1+kd \, ]$ :
\[\dfrac{k}{2}[ \, 2T_1+(k-1)d \, ]+[ \, T_1+kd \, ]=\dfrac{(k+1)}{2}[ \, 2T_1+kd \, ]\quad\checkmark\]
4
Step 4 — Conclusion
By Mathematical Induction, the arithmetic series sum formula holds for all integers $n\ge1$
✅ Answer
Proved — this is the arithmetic series formula $S_n=\dfrac{n}{2}[ \, 2T_1+(n-1)d \, ]$
4
Prove by mathematical induction that $T_1+T_1r+\c…
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5
Prove by mathematical induction that $1+6+15+\cdo…
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6
Prove by induction that $1\times2\times3+2\times3…
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7
Using mathematical induction, prove for all integ…
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8
Prove for all integers $n\ge1$ that $(n+1)(n+2)\c…
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9
If $U_1=1$ and $U_{k+1}=U_k+2k+1$, use induction …
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10
If $u_1=7$ and $u_n=2u_{n-1}-1$ for $n\ge2$, prov…
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11
The Fibonacci sequence: $T_1=1$, $T_2=2$, $T_n=T_…
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1
Solve $\begin{cases}3x+y=11\\x=3\end{cases}$ by the substitution method
▾
✦ Full Worked Solution
1
Step 1 — Substitute
Since $x=3$, substitute directly into the first equation:
\[3(3)+y=11\implies 9+y=11\implies y=2\]
2
Step 2 — State the Solution
\[x=3,\quad y=2\]
✅ Answer
$x=3,\ y=2$
2
Solve $\begin{cases}x+y=8\\x=5\end{cases}$ by the substitution method
▾
✦ Full Worked Solution
1
Step 1 — Substitute
Substitute $x=5$ into the first equation:
\[5+y=8\implies y=3\]
2
Step 2 — State the Solution
\[x=5,\quad y=3\]
✅ Answer
$x=5,\ y=3$
3
Solve $\begin{cases}3x+y=11\\x=y-3\end{cases}$ by the substitution method
▾
✦ Full Worked Solution
1
Step 1 — Substitute
Substitute $x=y-3$ into the first equation:
\[3(y-3)+y=11\implies 3y-9+y=11\implies 4y=20\implies y=5\]
2
Step 2 — Find x
\[x=5-3=2\]
3
Step 3 — State the Solution
\[x=2,\quad y=5\]
✅ Answer
$x=2,\ y=5$
4
Solve $\begin{cases}y=3x-2\\y=x+4\end{cases}$ by …
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5
Solve $\begin{cases}x=4y+2\\5x-y=28\end{cases}$ b…
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6
Solve $\begin{cases}y=x+4\\3x-2y+10=0\end{cases}$…
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7
Solve $\begin{cases}x+y=0\\-x+y=16\end{cases}$ by…
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8
Solve $\begin{cases}x-y=6\\\dfrac{x}{3}+\dfrac{y}…
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9
Solve $\begin{cases}y+x=9\\y-x=-3\end{cases}$ by …
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