Sum of n terms of 1 + 4 + 13 + 40 + ……………… isA) \(\cfrac{

1.	Sum of n terms of 1 + 4 + 13 + 40 + ……………… isA) \(\cfrac{(3^{n+1}+2n-3)}{4}\) B) \(\cfrac{(3^{n+1}-2n-3)}{4}\)C) \(\cfrac{(3^{n+1}-2n+3)}{4}\)D) \(\cfrac{(3^{n+1}+2n+3)}{4}\)
Answer» Correct option is (B) \(\frac{(3^{n+1}-2n-3)}{4}\) \(S_n=1+4+13+40+......+\) upto n terms \(=1+(1+3)+(1+3+9)+(1+3+9+27)+........+\) upto n terms \(=3^0+(3^0+3^1)+(3^0+3^1+3^2)\) \(+(3^0+3^1+3^2+3^3)+..........\) \(+(3^0+3^1+3^2+........+3^{n-1})\) \(=\sum(3^0+3^1+3^2+........+3^{n-1})\) \(=\sum\frac{1(3^n-1)}{3-1}\) \((\because\) Sum in G.P. \(=\frac{a(r^n-1)}{r-1})\) \(=\frac12(\sum 3^n-\sum1)\) \(=\frac12\left(\frac{3(3^n-1)}{3-1}-n\right)\) \(\left(\because\sum 3^n=\frac{3(3^n-1)}{3-1}\right)\) \(=\frac14(3^{n+1}-3-2n)\) \(=\frac{3^{n+1}-2n-3}{4}\) Correct option is B) \(\cfrac{(3^{n+1}-2n-3)}{4}\)

Sum of n terms of 1 + 4 + 13 + 40 + ……………… isA) \(\cfrac{(3^{n+1}+2n-3)}{4}\) B) \(\cfrac{(3^{n+1}-2n-3)}{4}\)C) \(\cfrac{(3^{n+1}-2n+3)}{4}\)D) \(\cfrac{(3^{n+1}+2n+3)}{4}\)

Answer»

Correct option is (B) \(\frac{(3^{n+1}-2n-3)}{4}\)

\(S_n=1+4+13+40+......+\) upto n terms

\(=1+(1+3)+(1+3+9)+(1+3+9+27)+........+\) upto n terms

\(=3^0+(3^0+3^1)+(3^0+3^1+3^2)\) \(+(3^0+3^1+3^2+3^3)+..........\) \(+(3^0+3^1+3^2+........+3^{n-1})\)

\(=\sum(3^0+3^1+3^2+........+3^{n-1})\)

\(=\sum\frac{1(3^n-1)}{3-1}\) \((\because\) Sum in G.P. \(=\frac{a(r^n-1)}{r-1})\)

\(=\frac12(\sum 3^n-\sum1)\)

\(=\frac12\left(\frac{3(3^n-1)}{3-1}-n\right)\) \(\left(\because\sum 3^n=\frac{3(3^n-1)}{3-1}\right)\)

\(=\frac14(3^{n+1}-3-2n)\)

\(=\frac{3^{n+1}-2n-3}{4}\)

Correct option is B) \(\cfrac{(3^{n+1}-2n-3)}{4}\)