The sum to n terms of the series 1 + 2 (1 + 1/n) + 3 (1 + 1/n)^2 + ......is

Question

The sum to n terms of the series 1 + 2 \(\big(1+\frac{1}{n}\big)\) + 3 \(\big(1+\frac{1}{n}\big)\)² + ......is

(a) n² 

(b) n (n – 1) 

(c) (n + 1)² 

(d) n (n + 1)

Answer 1

(a) n²

S_n = 1 + 2 \(\big(1+\frac{1}{n}\big)\) + 3 \(\big(1+\frac{1}{n}\big)\)² + ...... + n \(\big(1+\frac{1}{n}\big)\)^n-1

∴ \(\big(1+\frac{1}{n}\big)\)S_n = \(\big(1+\frac{1}{n}\big)\) + 2\(\big(1+\frac{1}{n}\big)\)² + ........+(n - 1)\(\big(1+\frac{1}{n}\big)\)^{n - 1} + n\(\big(1+\frac{1}{n}\big)\)ⁿ

\(\bigg(\because\text{is an A.G.P with common ratio}\,\big(1+\frac{1}{n}\big)\bigg)\)

⇒ S_n \(\bigg[\)1 - \(\big(1+\frac{1}{n}\big)\)\(\bigg]\) = 1 + \(\big(1+\frac{1}{n}\big)\) + \(\big(1+\frac{1}{n}\big)\)² + .... \(\big(1+\frac{1}{n}\big)\)^{n - 1} - n\(\big(1+\frac{1}{n}\big)\)ⁿ

⇒ \(-\frac{1}{n}\)S_n = \(\frac{1\bigg(\big(1+\frac{1}{n}\big)^n-1\bigg)}{\big(1+\frac{1}{n}\big)-1}\) - n\(\big(1+\frac{1}{n}\big)\)ⁿ

⇒ \(-\frac{1}{n}\)S_n = n\(\bigg[\)\(\big(1+\frac{1}{n}\big)\)n - 1\(\bigg]\) - n\(\big(1+\frac{1}{n}\big)\)ⁿ

⇒ \(-\frac{1}{n}\)S_n = - n ⇒ S_n = n².