Question

# Let $$A$$ be the sum of the first $$20$$ terms and $$B$$ be the sum of the first $$40$$ terms of the series $$1^{2} + 2.2^{2} + 3^{2} + 2.4^{2} + 5^{2} + 2.6^{2} + .....$$ If $$B - 2A = 100\lambda$$, then $$\lambda$$ is equal to

A
464
B
496
C
232
D
248

Solution

## The correct option is C $$248$$$$B=1^2+2.2^2+3^2+2.4^2+....+2.40^2$$$$A=1^2+2.2^2+3^2+2.4^2+.... +2.20^2$$$$B = 1^2+3^3+5^2+...+39^2+2 [2^2+4^2+...+40^2]$$$$A = 1^2+3^2+5^2+... + 19^2+2[2^2+4^2+...+20^2]$$Sum of square of first n odd natural number$$=\dfrac{n(2n+1)(2n-1)}{3}$$sum of square of first n even natural number$$=\dfrac{2n(n+1)(2n+1)}{3}$$$$B=\left[\dfrac{n(2n+1)(2n-1)}{3}\right]_{n=20} + 2\left[\dfrac{2n(n+1)(2n+1)}{3}\right]_{n=20}$$$$A=\left[\dfrac{n(2n+1)(2n-1)}{3}\right]_{n=10}+2\left[\dfrac{2n(n+1)2n+1)}{3}\right]_{n=10}$$$$B = \dfrac{20(41)(39)}{3} + 2\left[\dfrac{(400(21)(41)}{3}\right]$$$$B = \dfrac{100860}{3}$$$$A=\dfrac{10(21)(19)}{3}+2\left[\dfrac{20(11)(21)}{3}\right]$$$$A=\dfrac{13230}{3}$$$$B-2A = \left[\dfrac{100860-26460}{3}\right] = \dfrac{74400}{3}$$$$B-2A = 24800= 100\lambda$$$$\lambda = 248$$Mathematics

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