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Sir please give me solution of this question 
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Asked by sanjaysingh.srct | 25 Jul, 2019, 09:34: AM
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C o n s i d e r space t h e space s u m left parenthesis k plus 1 right parenthesis S subscript k plus fraction numerator k left parenthesis k plus 1 right parenthesis over denominator 1.2 end fraction S subscript k minus 1 end subscript plus fraction numerator left parenthesis k plus 1 right parenthesis k left parenthesis k minus 1 right parenthesis over denominator 1.2.3 end fraction S subscript k minus 2 end subscript plus.... plus left parenthesis k plus 1 right parenthesis S subscript 1 equals C presubscript 1 presuperscript k plus 1 end presuperscript space S subscript 1 plus C presubscript 2 presuperscript k plus 1 end presuperscript space S subscript 2 plus.... C presubscript k presuperscript k plus 1 end presuperscript space S subscript k equals C presubscript 1 presuperscript k plus 1 end presuperscript space open parentheses 1 plus 2 plus 3 plus.. plus n close parentheses plus C presubscript 2 presuperscript k plus 1 end presuperscript space open parentheses 1 squared plus 2 squared plus... plus n squared close parentheses plus.... C presubscript k presuperscript k plus 1 end presuperscript space open parentheses 1 to the power of k plus 2 to the power of k plus... plus n to the power of k close parentheses equals open parentheses C presubscript 1 presuperscript k plus 1 end presuperscript space cross times space 1 space plus space C presubscript 2 presuperscript k plus 1 end presuperscript space cross times space 1 squared space plus space C presubscript k presuperscript k plus 1 end presuperscript space 1 to the power of k close parentheses plus open parentheses C presubscript 1 presuperscript k plus 1 end presuperscript space cross times space 2 space plus space C presubscript 2 presuperscript k plus 1 end presuperscript space cross times space 2 squared space plus space C presubscript k presuperscript k plus 1 end presuperscript space 2 to the power of k close parentheses space plus... plus open parentheses C presubscript 1 presuperscript k plus 1 end presuperscript space cross times space n space plus space C presubscript 2 presuperscript k plus 1 end presuperscript space cross times space n squared space plus space C presubscript k presuperscript k plus 1 end presuperscript space n to the power of k close parentheses equals open square brackets open parentheses 1 plus 1 close parentheses to the power of k plus 1 end exponent minus 1 minus C presubscript k plus 1 end presubscript presuperscript k plus 1 end presuperscript space 1 to the power of k plus 1 end exponent close square brackets plus open square brackets open parentheses 1 plus 2 close parentheses to the power of k plus 1 end exponent minus 1 minus C presubscript k plus 1 end presubscript presuperscript k plus 1 end presuperscript space 2 to the power of k plus 1 end exponent close square brackets plus... plus open square brackets open parentheses 1 plus n close parentheses to the power of k plus 1 end exponent minus 1 minus C presubscript k plus 1 end presubscript presuperscript k plus 1 end presuperscript space n to the power of k plus 1 end exponent close square brackets equals open square brackets 2 to the power of k plus 1 end exponent minus 1 minus space 1 to the power of k plus 1 end exponent close square brackets plus open square brackets 3 to the power of k plus 1 end exponent minus 1 minus space 2 to the power of k plus 1 end exponent close square brackets plus... plus open square brackets open parentheses n plus 1 close parentheses to the power of k plus 1 end exponent minus 1 minus space n to the power of k plus 1 end exponent close square brackets equals open parentheses n plus 1 close parentheses to the power of k plus 1 end exponent minus 1 to the power of k plus 1 end exponent minus n S i n c e comma space S subscript 0 equals n T h e r e f o r e comma S subscript 0 plus C presubscript 1 presuperscript k plus 1 end presuperscript space S subscript 1 plus C presubscript 2 presuperscript k plus 1 end presuperscript space S subscript 2 plus.... C presubscript k presuperscript k plus 1 end presuperscript space S subscript k equals n plus open parentheses n plus 1 close parentheses to the power of k plus 1 end exponent minus 1 minus n equals open parentheses n plus 1 close parentheses to the power of k plus 1 end exponent minus 1 H e n c e comma space left parenthesis k plus 1 right parenthesis S subscript k plus fraction numerator k left parenthesis k plus 1 right parenthesis over denominator 1.2 end fraction S subscript k minus 1 end subscript plus fraction numerator left parenthesis k plus 1 right parenthesis k left parenthesis k minus 1 right parenthesis over denominator 1.2.3 end fraction S subscript k minus 2 end subscript plus.... plus left parenthesis k plus 1 right parenthesis S subscript 1 plus S subscript 0 equals open parentheses n plus 1 close parentheses to the power of k plus 1 end exponent minus 1
Answered by Renu Varma | 25 Jul, 2019, 10:48: AM

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