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If n is a positive integer satisfying the equation `2+(6*2^(2)-4*2)+(6*3^(2)-4*3)+"......."+(6*n^(2)-4*n)=140` then the value of n is |
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Answer» `:.2+(6*2^(2)-4*2)+(6*3^(2)-4*3)+"......."+(6*n^(2)-4*n)=140` `2+6(2^(2)+3^(2)+"......."+n^(2))-4*(2+3+"....."+n)=140` `implies 2+6((n(n+1)(2n-1))/(6)-1)-4((n(n+1))/(2)-1)=140` `implies 2+n(n+1)(2n+1)-6-2n(n+1)+4=140` `implies n(n+1)(2n+1)-2n(n+1)-140=0` `implies 2n^(3)3n^(2)+n-2n^(2)-2n-140=0` `implies 2n^(3)+n^(2)-n-140=0` `implies (n-4)+(2n^(2)+9n+35)=0` `implies n=4 " or " 2n^(2)+9n+35=0` `implies 2n^(2)+9n+35=0` `implies n=(-9pmsqrt(81-280))/(4)` `:. n=(9pmsqrt(-199))/(4) " " [" complex values "]` Only positive integer value of n is 4. |
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