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Vidaara.orgClass 11 · Mathematics
CodeVID-M11-15-SUM-01
Summation & Divisibility by Induction — Assignment
Chapter: Principle of Mathematical Induction
Topic: Proving Summation and Divisibility Statements
Maximum Marks: 35
Time: 75 minutes
Name: ____________________ Roll No.: __________ Date: ____________

General Instructions

  • All questions are compulsory.
  • Section A carries 1 mark each, Section B 2 marks, Section C 3 marks and Section D 5 marks.
  • Show all working for Sections B, C and D. Only final answers are given at the end — for full solutions, raise your doubts with your teacher.
Section A — Multiple Choice Questions 5 × 1 = 5 marks
1.
$1+2+\cdots+n$ equals:
  • A.$n^2$
  • B.$\dfrac{n(n+1)}{2}$
  • C.$\dfrac{n(n+1)(2n+1)}{6}$
  • D.$2^n-1$
2.
$1^3+2^3+\cdots+n^3$ equals:
  • A.$\left[\dfrac{n(n+1)}{2}\right]^2$
  • B.$\dfrac{n(n+1)}{2}$
  • C.$n^3$
  • D.$\dfrac{n(n+1)(2n+1)}{6}$
3.
$n^3-n$ is divisible by:
  • A.$2$
  • B.$4$
  • C.$6$
  • D.$5$
4.
$7^n-3^n$ is divisible by:
  • A.$2$
  • B.$3$
  • C.$4$
  • D.$7$
5.
$2+4+\cdots+2n$ equals:
  • A.$n^2$
  • B.$n(n+1)$
  • C.$2n$
  • D.$\dfrac{n(n+1)}{2}$
Section B — Short Answer (2 marks) 4 × 2 = 8 marks
6.
Verify the base case of $1+2+\cdots+n=\dfrac{n(n+1)}{2}$.
7.
Write $f(k+1)$ for $f(n)=7^n-3^n$ in terms of $7^k$ and $3^k$.
8.
State the sum of the first $n$ squares.
9.
For $f(n)=3^{2n}-1$, compute $f(1)$ and its divisor.
Section C — Short Answer (3 marks) 4 × 3 = 12 marks
10.
Simplify $\dfrac{k(k+1)(2k+1)}{6}+(k+1)^2$ and factor.
11.
Show $(k+1)^3-(k+1)=(k^3-k)+3k(k+1)$.
12.
Write $3^{2(k+1)}-1$ using the hypothesis $3^{2k}=8m+1$.
13.
Prove the inductive step for $2+4+\cdots+2n=n(n+1)$.
Section D — Long Answer (5 marks) 2 × 5 = 10 marks
14.
Prove by induction that $1^2+2^2+\cdots+n^2=\dfrac{n(n+1)(2n+1)}{6}$.
15.
Prove by induction that $7^n-3^n$ is divisible by $4$ for all $n\in\mathbb{N}$.

Answer Key

Section A — Multiple Choice Questions
  1. (B) $\dfrac{n(n+1)}{2}$
  2. (A) $\left[\dfrac{n(n+1)}{2}\right]^2$
  3. (C) $6$
  4. (C) $4$
  5. (B) $n(n+1)$
Section B — Short Answer (2 marks)
  1. $n=1$: $1=\dfrac{1\cdot2}{2}=1$; true.
  2. $f(k+1)=7\cdot7^k-3\cdot3^k$.
  3. $\dfrac{n(n+1)(2n+1)}{6}$.
  4. $f(1)=8$, divisible by $8$.
Section C — Short Answer (3 marks)
  1. $\dfrac{(k+1)(k+2)(2k+3)}{6}$.
  2. Expand $(k+1)^3-(k+1)=k^3+3k^2+3k-k=(k^3-k)+3k^2+3k=(k^3-k)+3k(k+1)$.
  3. $9(8m+1)-1=72m+8=8(9m+1)$.
  4. $k(k+1)+2(k+1)=(k+1)(k+2)$.
Section D — Long Answer (5 marks)
  1. Base $n=1$ true; assuming the $k$-case and adding $(k+1)^2$ gives $\dfrac{(k+1)(k+2)(2k+3)}{6}$; by PMI it holds for all $n$.
  2. Base $n=1$: $4$; step $7^{k+1}-3^{k+1}=7(7^k-3^k)+4\cdot3^k=4(7m+3^k)$; by PMI $4\mid(7^n-3^n)$.
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