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Find the sum of n terms of the series (a+b)+(a^(2)+ab+b^(2))+(a^(3)+a^(2)b+ab^(2)+b^(3))+"......." where a ne 1,bne 1 and a ne b. |
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Answer» Solution : ` a ne1 B ne 0 and ane b ` Let S = ( a+ b) + `(a^(2) + ab + b^(2)+ ( a^(3) + a^(2) b+ ab^(2) +b^(3) + ….+ n ` terms `=(1) /( (a-b) ) [(a^(2)-b^(2)) + ( a^(3) -b ^(3)) + (a^(4) -b^(4) ) + ....+_n terms ]` `= (1) /((a-b) ) [a^(2) (1+ a+ ......+ nterms )- b ^(2)(1+ b+ b^(2) + .....+nterms ) ` ` =(1)/( (a- b) ) [ a^(2)* (1* (a^(n) -1) )/( (a-1) ) -b^(2) ( 1* (b^(n) -1))/( (b-1) ) ]` ` "" = (1)/( (a-b ) ) [a^(2) ((1- n^(2)))/( (1- a) ) - b^(2) ( ( 1- b^(n)))/( (1-b) ) ]` |
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