Q. 305.0( 1 Vote )

Evaluate the foll

Answer :


Formula used:



where,



Here, a = 1 and b = 3


Therefore,




Let,



Here, f(x) = x2 + x and a = 1



Now, by putting x = 1 in f(x) we get,


f(1) = 12 + 1 = 1 + 1 = 2


f(1 + h)


= (1 + h)2 + (1 + h)


= h2 + 12 + 2(h)(1) + 1 + h


= h2 + 2h + h + 1 + 1


= h2 + 3h + 2


Similarly, f(1 + 2h)


= (1 + 2h)2 + (1 + 2h)


= (2h)2 + 12 + 2(2h)(1) + 1 + 2h


= (2h)2 + 4h + 2h + 1 + 1


= (2h)2 + 6h + 2


{ (x + y)2 = x2 + y2 + 2xy}




In this series, 2 is getting added n times



Now take h2 and 3h common in remaining series





Put,



Since,
















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