The least value of n so that 1 + 32 + 33 + 3 + ………….. n terms

1.	The least value of n so that 1 + 32 + 33 + 3 + ………….. n terms is greater than 2000 is ……………. A) 8B) 7 C) 9 D) 6
Answer» Correct option is (A) 8 \(\because\) \(1+3+3^2+3^3+.....\) will form a G.P. with common ratio 3 whose sum of n terms is given by \(\frac{a(r^n-1)}{r-1}.\) \(S_n=1+3+3^2+3^3+......+n\) terms \(=\frac{1(3^n-1)}{3-1}=\frac{3^n-1}2\) Let \(S_n>2000\) \(\Rightarrow\) \(\frac{3^n-1}2>2000\) \(\Rightarrow3^n-1>4000\) \(\Rightarrow3^n>4001\) \(\because3^7=2187<4001\) But \(3^8=6561>4001\) Hence, least value of n is 8 so that the sum \(1+3+3^2+3^3+.....\) is greater than 2000. Correct option is A) 8

The least value of n so that 1 + 32 + 33 + 3 + ………….. n terms is greater than 2000 is ……………. A) 8B) 7 C) 9 D) 6

Answer»

Correct option is (A) 8

\(\because\) \(1+3+3^2+3^3+.....\) will form a G.P. with common ratio 3 whose sum of n terms is given by \(\frac{a(r^n-1)}{r-1}.\)

\(S_n=1+3+3^2+3^3+......+n\) terms

\(=\frac{1(3^n-1)}{3-1}=\frac{3^n-1}2\)

Let \(S_n>2000\)

\(\Rightarrow\) \(\frac{3^n-1}2>2000\)

\(\Rightarrow3^n-1>4000\)

\(\Rightarrow3^n>4001\)

\(\because3^7=2187<4001\)

But \(3^8=6561>4001\)

Hence, least value of n is 8 so that the sum \(1+3+3^2+3^3+.....\) is greater than 2000.

Correct option is A) 8