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  • Find the sum of 10 terms of the series   <div class='qna-opti

Find the sum of 10 terms of the series 1+3+7+13+21+....

 

Option: 1

100


Option: 2

300


Option: 3

340


Option: 4

None of these


Answers (1)

best_answer

Difference of consecutive terms are 2,4,6,8,.... which is in AP.

So, assuming a_n = an^2+bn +c

Put n = 1 : a_1 = a+b +c\Rightarrow 1 =a+b+c                ........(1)

Put n = 2: a_2 = 4a+2b +c\Rightarrow 3 =4a+2b+c               .........(2)

Put n = 3: a_3 = 9a+3b +c\Rightarrow 7 =9a+3b+c                ...........(3)

Now to get a , b, c

(3) - (2) \Rightarrow 5a +b = 4                    ........(4)

(2) - (1) \Rightarrow 3a +b = 2                     .........(5)

(4) - (5) \Rightarrow 2a = 2\Rightarrow a=1

Using (5) \Rightarrow b=-1

Using (1) \Rightarrow c=1

So a_n = n^2-n+1

\sum a_n=\sum n^2-\sum n+\sum 1=\frac{n(n+1)(2n+1)}{6}-\frac{n(n+1)}{2}+n

So S_{10}=\frac{10.11.21}{6}-\frac{10.11}{2}+10=340

Posted by

Riya

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