nguyendinhtuong wrote:

What is the value of A?

\(A=1 \times 2+2 \times 3+...+9 \times 10\)

A. 198

B. 240

C. 330

D. 440

E. 572

So the nth term is given by

\(T_n = n * (n + 1)\)

\(T_n = n^2 + n\)

So the first term is equivalent to 1^2 + 1

The second term is equivalent to 2^2 + 2

The third term is equivalent to 3^2 + 3

...

If we add the 9 terms, we get: \(1^2 + 1 + 2^2 + 2 + 3^2 + 3 + ... 9^2 + 9\)

Note that we know how to handle these two:

1 + 2 + 3 ... + 9

This is given by n(n+1)/2 = 9*(10)/2 = 45

and

1^2 + 2^2 + 3^2 + ... + 9^2

This is given by n(n+1)(2n+1)/6 = 9*10*19/6 = 285

Total sum = 45 + 285 = 330

Answer (C)

What is the source of this question? I doubt it is a GMAT specific source.

_________________

Karishma

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