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CBSE Class 11-commerce Answered

If sk=1+2+3+....k÷k.Find the value of s1^2+s2^2+s3^2+....sk^2
Asked by Saumil.sharma2 | 30 Nov, 2018, 11:00: PM
answered-by-expert Expert Answer
it is given in the posting sk = 1+2+3......k÷k . This I assume as begin mathsize 12px style s subscript k equals space fraction numerator 1 plus 2 plus 3.... plus k over denominator k end fraction end style  .  
With this assumption, this question is solved as given below

begin mathsize 12px style S space equals space s subscript 1 superscript 2 space plus space s subscript 2 superscript 2 space plus space s subscript 3 superscript 2 space............. s subscript k superscript 2 space equals space 1 squared space plus space open parentheses fraction numerator 1 plus 2 over denominator 2 end fraction close parentheses squared space plus space open parentheses fraction numerator 1 plus 2 plus 3 over denominator 3 end fraction close parentheses squared space plus.............. open parentheses fraction numerator 1 plus 2 plus 3.... k over denominator k end fraction close parentheses squared
E a c h space t e r m space c o n t a i n s space w i t h i n space t h e space b r a c k e t space s u m space o f space n space n a t u r a l space n u m b e r s space d i v i d e d space b y space n

n to the power of t h end exponent space t e r m space equals t subscript n space equals space open parentheses fraction numerator n open parentheses n plus 1 close parentheses over denominator 2 cross times n end fraction close parentheses squared space equals space 1 fourth open parentheses n squared plus 2 n plus 1 close parentheses
H e n c e space S space equals space 1 fourth sum from n equals 1 to k of left parenthesis n squared plus 2 n plus 1 right parenthesis space equals space 1 fourth sum from n equals 1 to k of n squared space plus space stack 1 half sum with n equals 1 below and k on top n space plus space k over 4 space equals space fraction numerator k open parentheses k plus 1 close parentheses open parentheses 2 k plus 1 close parentheses over denominator 24 end fraction plus fraction numerator k open parentheses k plus 1 close parentheses over denominator 4 end fraction plus k over 4

S space space equals space k over 24 open parentheses 2 k squared plus 9 k plus 13 close parentheses end style
Answered by Thiyagarajan K | 01 Dec, 2018, 07:46: AM

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