If $S = 1 + 2 + 4 + 8 + 16 + 32 + . . . . . \infty$ . then $S$ is a +ive number. Multiply both sides by $2$ , then it is found that $2 S = S - 1$ which leads to conclusion $S = - 1$ which is certainly negative. Do you agree with the conclusion ? If you agree with the conclusion write 1 otherwise 0

Open in App

Solution

Since here common ratio is 2 which is greater than 1, hence S is infinite so the first step i.e. $2 S = S - 1$ is correct but conclusion from this that $S = - 1$ is false since we cannot cancel S i.e $\infty$ from both sides.

Suggest Corrections

Similar questions

S \leq P \leq A = R

S \leq E

R < O \leq L \leq E

G = E \geq S

P \leq S

P \leq E

2 \times 10^{- 3}

l^{- 1} S^{- 1}

\to

(+ \frac{1}{2}, 0 and - \frac{1}{2})

Statements: P $$$ Q, Q $δ$ S, S * R

Conclusions:

I. S * P

II. S $$$ P

Join BYJU'S Learning Program