Grade 12th passAlgebra

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The sum of n terms of the following series; 13 + 33 + 53 + 73 + .... is
(A) n2(2n2 – 1)
(B) n3(n – 1)
(C) n3 + 8n + 4
(D) 2n4 + 3n2

sohan goswami

8 Years agoGrade 12th pass

4 Answers

Arun

8 Years ago

Dear Sohan

Please uae hit and trial. It will save your time.

As put n = 1

Hence sum should be equal to 13. Which is option C

Samyak Jain

8 Years ago

Let S denote the sum of n terms of the series T_nbe the n^th term of the series.

S = 13 + 33 + 53 + 73 + ....... + T_{n ..................(1)}

S = 13 + 33 + 53 + ....... + T_{n –1}+ T_{n ….............(2)}

Subtract (2) from (1)

0 = 13 + (20 + 20 + 20 + …........ n – 1 times) – T_n

$\therefore$ T_n = 13 + (n – 1) * 20 = 20n – 20 + 13 = 20n – 7

$\therefore$ S = $\sum_{}^{}$ _n=1ⁿ⁼ⁿ T_n = ∑ 20n – ∑7

= [20n(n + 1) / 2] – 7n

= 10n(n+1) – 7n

= 10n²+ 10n – 7n

= 10n²+ 3n

S = n (10n + 3) is the correct answer.

All the options given are incorrect !!!!

Option C is valid only for n=1.

Govind

8 Years ago

Please uae hit and trial. It will save your time.As put n = 1Hence sum should be equal to 13. Which is option C orLet S denote the sum of n terms of the series Tn be the nth term of the series.S = 13 + 33 + 53 + 73 + ....... + Tn ..................(1)S = 13 + 33 + 53 + ....... + Tn –1 + Tn ….............(2)Subtract (2) from (1)0 = 13 + (20 + 20 + 20 + …........ n – 1 times) – Tn\therefore Tn = 13 + (n – 1) * 20 = 20n – 20 + 13 = 20n – 7\therefore S = \sum_{}^{}n=1n=n Tn = ∑ 20n – ∑7 = [20n(n + 1) / 2] – 7n = 10n(n+1) – 7n = 10n2 + 10n – 7n = 10n2 + 3nS = n (10n + 3) is the correct answer.All the options given are incorrect !!!!Option C is valid only for n=1.