What is the value of A=1*2+2*3+...+9*10?

Question

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

KarishmaB · Accepted Answer

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.

parmeshwarsharma · Answer

N*(N+1)*(N+2)/3

Put N=9

We get 9*10*11/3=30*11=330

Sent from my ONE A2003 using  GMAT Club Forum mobile app

Abhishek009 · Answer

No way to deny  parmeshwarsharma  's is the best way to solve this question...

A = ( 1*2 ) + (2*3) + ( 3*4) + (4*5) + (5*6) + (6*7) + ( 7*8 ) + ( 8*9 ) + (9*10)

A = 2 + 6 + 12 + 20 + 30 + 42 + 56 + 72 + 90

A = ( 2 + 6 + 12 + 20 + 30 ) + ( 42 + 56 + 72 + 90 )

A = 70 + 260

A = 330

So, Answer will be (C)

stonecold · Answer

Hi i did it the way you did.
I mean i tried using the using digit concept but since it was zero i found the value instead.
What did  parmeshwarsharma  do there?

Did he use the concept of summation?
As in sum of n terms is n*n+1/2 or something.
You mind explaining it a bit

Regards
Stone Cold

Abhishek009 · Answer

t_1 = 1^2 + 1 
 t_2 = 2^2 + 2 
____________
 t_n = n^2 + n

S_n = ( t_1 + t_2.......t_n ) + ( 1 + 2 ....n )

S_n = \frac{n ( n + 1 ) ( 2n + 1 )}{6} + \frac{n ( n +1)}{2}

Solve it you will get

S_n = \frac{n( n+1 )( n+2 )}{3}

I highly recommend not to go into such details , GMAT doesn't want you to recognize it, if you are not familiar with the same.. Try to solve the method I have used if you are not familiar with the concept...

sjavvadi · Answer

Please Kudos+1 if you like the solution

broall · Answer

Here is the solution for this one.

$3k(k+1)=k(k+1) =-(k-1)k(k+1)+k(k+1)(k+2)$

We have $3A=3\times 1 \times 2 + 3 \times 2 \times 3 +...+ 3 \times 9 \times 10$

Hence
$\begin{split}
3\times 1 \times 2 &= - 0\times 1 \times 2 + 1 \times 2 \times 3\
3\times 2 \times 3 &= - 1\times 2 \times 3 + 2 \times 3 \times 4\
&...\
3\times 9 \times 10 &= - 8\times 9 \times 10 + 9 \times 10 \times 11\
3\times 1 \times 2 + 3 \times 2 \times 3 +...+ 3 \times 9 \times 10 &= - 0\times 1 \times 2 + 9 \times 10 \times 11 \
3A &= 9 \times 10 \times 11\
A&= \frac{ 9 \times 10 \times 11}{3}
\end{split}$

Nunuboy1994 · Answer

What exactly is this formula?

KrishnakumarKA1 · Answer

we can say that ant term can be expressed as N(N+1) = N^2 +N

summing up from 1 to 10 we get N(N+1)(2N+1)/6 + N(N+1)/2 . Put n = 10, we get the sum as 330

bumpbot · Answer

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