Question

For all $n\in N$, consider the following statements
I. $\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\cdots +\frac{1}{2^n}=1-\frac{1}{2^n}$
II. $\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\cdots +\frac{1}{2^n}=1+\frac{1}{2^n}$
III. $\frac{1}{2\cdot 5}+\frac{1}{5\cdot 8}+\frac{1}{8\cdot 11}+\cdots +\frac{1}{(3n-1)(3n+2)}=\frac{n+1}{6n+4}$
IV. $\frac{1}{2\cdot 5}+\frac{1}{5\cdot 8}+\frac{1}{8\cdot 11}+\cdots +\frac{1}{(3n-1)(3n+2)}=\frac{n}{6n+4}$
Choose the correct option.

• A. II and III are not true
• B. Only II is not true
• C. Only IV is not true
• D. I and IV are not true

Answer 1

Answer: (A)

I. Let the statement $P(n)$ be defined as
i.e., $P(n):\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\dots +\frac{1}{2^n}=1-\frac{1}{2^n}$
Step I For $n=1$,
$P(1):\frac{1}{2}=1-\frac{1}{2^1}=1-\frac{1}{2}=\frac{1}{2}$
which is true.
Step II Let it is true for $n=k$,
i.e., $\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\dots +\frac{1}{2^k}=1-\frac{1}{2^k}.....$(i)
Step III For $n=k+1$,
$\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\dots +\frac{1}{2^k}\right)+\frac{1}{2^{k+1}}=1-\frac{1}{2^k}+\frac{1}{2^{k+1}}$
[using Eq. (i)]
$=1-\frac{1}{2^k}+\frac{1}{2^k\cdot 2}$
$=1-\frac{1}{2^k}\left(1-\frac{1}{2}\right)($ taking $\frac{1}{2^k}$ common in last two terms $)$
$=1-\frac{1}{2^k}\times \frac{1}{2}=1-\frac{1}{2^{k+1}}$
Therefore, $P(k+1)$ is true when $P(k)$ is true. Hence. from the principle of mathematical induction, the statement is true for all natural numbers $n$.
So, $I$ is true.
IV. Let the statement $P(n)$ be defined as
i.e., $P(n):\frac{1}{2\cdot 5}+\frac{1}{5\cdot 8}+\frac{1}{8\cdot 11}+\dots +\frac{1}{(3n-1)(3n+2)}$
$=\frac{n}{6n+4}$
Step I For $n=1$,
$P(1):\frac{1}{2\cdot 5}=\frac{1}{6\times 1+4}=\frac{1}{10}=\frac{1}{2\cdot 5}$
which is true.
Step II Let it is true for $n=k$,
i.e., $\frac{1}{2\cdot 5}+\frac{1}{5\cdot 8}+\frac{1}{8\cdot 11}+\dots +\frac{1}{(3k-1)(3k+2)}$
$=\frac{k}{6k+4}....$(i)
Step III For $n=k+1$,
$[\frac{1}{2\cdot 5}+\frac{1}{5\cdot 8}+\frac{1}{8\cdot 11}+\dots +\frac{1}{(3k-1)(3k+2)}]$
$=\frac{k}{6k+4}+\frac{1}{(3k+2)(3k+5)}\text{ [using Eq. (i)] }$
$-\frac{k}{2(3k+2)}+\frac{1}{(3k+2)(3k+5)}$
$=\frac{k(3k+5)+2}{2(3k+2)(3k+5)}=\frac{3k^2+5k+2}{2(3k+2)(3k+5)}$
$=\frac{3k^2+3k+2k+2}{2(3k+2)(3k+5)}=\frac{3k(k+1)+2(k+1)}{2(3k+2)(3k+5)}$
$=\frac{(k+1)(3k+2)}{2(3k+2)(3k+5)}$
$=\frac{k+1}{2(3k+3+2)}=\frac{k+1}{2[3(k+1)+2]}$
$=\frac{k+1}{6(k+1)+4}$
Therefore, $P(k+1)$ is true when $P(k)$ is true. Hence, from the principle of mathematical induction, the statement is true for all natural numbers $n$.
So, IV is true.
Hence, II and III are not true.