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. $\frac{1}{2 \cdot 5}+\frac{1}{5 \cdot 8}+\frac{1}{8 \cdot 11}+\cdots+\frac{1}{(3 n-1)(3 n+2)}=\frac{n+1}{6 n+4}$
IV. $\frac{1}{2 \cdot 5}+\frac{1}{5 \cdot 8}+\frac{1}{8 \cdot 11}+\cdots+\frac{1}{(3 n-1)(3 n+2)}=\frac{n}{6 n+4}$
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., $P(n): \frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\ldots+\frac{1}{2^n}=1-\frac{1}{2^n}$
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}+\ldots+\frac{1}{2^k}=1-\frac{1}{2^k} .....$(i)
Step III For $n=k+1$,
$\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\ldots+\frac{1}{2^k}\right)+\frac{1}{2^{k+1}}=1-\frac{1}{2^k}+\frac{1}{2^{k+1}}$
[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}+\ldots+\frac{1}{(3 n-1)(3 n+2)} $
$=\frac{n}{6 n+4}$
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., $\frac{1}{2 \cdot 5}+\frac{1}{5 \cdot 8}+ \frac{1}{8 \cdot 11}+\ldots+\frac{1}{(3 k-1)(3 k+2)}$
$ =\frac{k}{6 k+4} ....$(i)
Step III For $n = k+1$,
${\left[\frac{1}{2 \cdot 5}\right.} \left.+\frac{1}{5 \cdot 8}+\frac{1}{8 \cdot 11}+\ldots+\frac{1}{(3 k-1)(3 k+2)}\right] $
$=\frac{k}{6 k+4}+\frac{1}{(3 k+2)(3 k+5)} \text { [using Eq. (i)] } $
$ -\frac{k}{2(3 k+2)}+\frac{1}{(3 k+2)(3 k+5)} $
$ =\frac{k(3 k+5)+2}{2(3 k+2)(3 k+5)}=\frac{3 k^2+5 k+2}{2(3 k+2)(3 k+5)} $
$=\frac{3 k^2+3 k+2 k+2}{2(3 k+2)(3 k+5)}=\frac{3 k(k+1)+2(k+1)}{2(3 k+2)(3 k+5)} $
$ =\frac{(k+1)(3 k+2)}{2(3 k+2)(3 k+5)} $
$=\frac{k+1}{2(3 k+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.