Answer :


Formula used:



where,



Here, a = 2 and b = 3


Therefore,




Let,



Here, f(x) = 2x2 + 1 and a = 2



Now, by putting x = 2 in f(x) we get,


f(2) = 2(22) + 1 = 2(4) + 1 = 8 + 1 = 9


f(1 + h)


= 2(2 + h)2 + 1


= 2{h2 + 22 + 2(h)(2)} + 1


= 2(h)2 + 8 + 2(4h) + 1


= 2(h)2 + 9 + 8(h)


Similarly, f(2 + 2h)


= 2(2 + 2h)2 + 1


= 2{2(2h)2 + 22 + 2(2h)(2)} + 1


= 2(2h)2 + 8 + 8(2h) + 1


= 2(2h)2 + 9 + 8(2h)


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




In this series, 9 is getting added n times



Now take 2h2 and 4h common in remaining series





Put,



Since,
















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