Question 72

Let $$a_1, a_2, ...$$ be integers such that
$$a_1 - a_2 + a_3 - a_4 + .... + (-1)^{n - 1} a_n = n,$$ for all $$n \geq 1.$$
Then $$a_{51} + a_{52} + .... + a_{1023}$$ equals

Solution

$$a_1 - a_2 + a_3 - a_4 + .... + (-1)^{n - 1} a_n = n$$

It is clear from the above equation that when n is odd, the co-efficient of a is positive otherwise negative.

$$a_1 - a_2 = 2$$

$$a_1 = a_2 + 2$$   

$$a_1 - a_2 + a_3 = 3$$

On substituting the value of $$a_1$$ in the above equation, we get

$$a_3$$ = 1

$$a_1 - a_2 + a_3 - a_4 = 4$$

On substituting the values of $$a_1, a_3$$ in the above equation, we get

$$a_4$$ = -1

$$a_1 - a_2 + a_3 - a_4 +a_5 = 5$$

On substituting the values of $$a_1, a_3, a_4$$ in the above equation, we get

$$a_5$$ = 1

So we can conclude that $$a_3, a_5, a_7....a_{n+1}$$ = 1 and $$a_2, a_4, a_6....a_{2n}$$ = -1

Now we have to find the value of $$a_{51} + a_{52} + .... + a_{1023}$$

Number of terms = 1023=51+(n-1)1

n=973

There will be 486 even and 487 odd terms, so the value of $$a_{51} + a_{52} + .... + a_{1023}$$ = 486*-1+487*1=1

Video Solution

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