K=1*2+2*3+..................+21*22

Solution

Given:
• K = 1 * 2 + 2 * 3 + 3 * 4 + … + 21 * 22

To find:
• The remainder when K is divided by 23

Approach and Working:
If we observe the terms of the series K, we can generalise it as follows:

• 1st term = 1 * 2
• 2nd term = 2 * 3
• 3rd term = 3 * 4
• Hence, nth term = n * (n + 1) = \(n^2 + n\)

Now, sum of all the terms of K = \(∑ n^2 + ∑ n\) = \(\frac{{n (n + 1) (2n + 1)}}{6} + \frac{n (n + 1)}{2}\)
Simplifying, we get K = \(\frac{{n (n + 1) (n + 2)}}{3}\)

• If n = 21, K = \(\frac{21 * 22 * 23}{3}\)

o Therefore, it will be always divisible by 23, giving a 0 remainder

Hence, the correct answer is option E.

Answer: E
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EgmatQuantExpert

Can we approach the question in the following manner:

Given K = 1*2 + 2*3 + 3*4+.......+ 21*22

Consider P = 1*2 + 2*3 + 3*4+......+ 21*22 + 22*23

P - K = 22*23 = a Multiple of 23

Hence P & K are both divisible by 23. Therefore when K is divided by 23, the remainder is 0.

Thanks,
GyM
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How do you derive the formula for sum of all terms of k?

Hey jenks88,
We didn't derive the formula for sum of all terms of k. We figured out the general expression representing any term of the series k (which is \(n^2 + n\)).

Once we have the general term, we applied the formula to find the sum of those elements forming the general term, to get the sum of all elements of k.

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

\(∑n = \frac{n (n + 1)}{2}\)

Hope this answers your query.
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I get the general term equation but still unsure where ∑\(n^{2}\)=\(\frac{n(n+1)(2n+1)}{6}\) comes from?

Hey jenks88 ,

I would never prefer to bother about how this formula is derived. I can directly use this formula whenever required.

But if you are still curious to know how we arrived at this formula, you can refer here : https://www.math-only-math.com/sum-of-th ... mbers.html
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Thanks to abhimahna. Please make a note of the most used 3 cases as following:

• Sum of first n natural numbers = \(∑n = 1 + 2 + 3 + ... + n = \frac{n(n+1)}{2}\)

• Sum of first n square numbers = \(∑n^2 = 1^2 + 2^2 + 3^2 + ... + n^2 = \frac{n(n+1)(2n+1)}{6}\)

• Sum of first n cube numbers = \(∑n^3 = 1^3 + 2^3 + 3^3 + ... + n^3 = [\frac{n(n+1)}{2}]^2\)

Also note that, you can use them only when the numbers are consecutive and start from 1

Hope it helps
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Hi Jenks88,

I would suggest to memorize the below formulae :-

Arithmetic progression:-

1. 1 + 2 + 3 + ...... + n = [n(n + 1)]/2.

2.1² + 2² + 3² +……………. + n² = [n(n+ 1)(2n+ 1)]/6.

3.1³ + 2³ + 3³ + . . . . + n³ = [{n(n + 1)}/2 ]

Geometric Progression:-

(i) The general form of a G.P. is a, ar, ar², ar³, . . . . . where a is the first term and r, the common ratio of the G.P.

(ii) The n th term of the above G.P. is t₀ = a.rn−1 .

(iii) The sum of first n terms of the above G.P. is S = a ∙ [(1 - rⁿ)/(1 – r)] when -1 < r < 1

or, S = a ∙ [(rⁿ – 1)/(r – 1) ]when r > 1 or r < -1.

(iv) The geometric mean of two positive numbers a and b is √(ab) or, -√(ab).

(v) a + ar + ar² + ……………. ∞ = a/(1 – r) where (-1 < r < 1).

Also, practice several sums on summation & series so that you would be able to master the application of the above formulae.

Hope it helps.
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Hey GyMrAT,
I have a query regarding the highlighted portion.
If we take a similar example, let's say
17 - 12 is a multiple of 5, does that necessarily mean both 17 and 12 are divisible by 5?
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*FacePalm*, damn i am smoking some weird ****, i guess.

Thanks EgmatQuantExpert, for kicking my ass.
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Beside solution of EgmatQuantExpert but you have to memorize the formula of the sum of consecutive squares, I want to provide the solution without memorizing that formula:

\(S = 1 \times 2 + 2 \times 3 + {...} + (n-1) \times n\)

Note that \(3 \times k \times (k+1) = [(k+2) - (k-1)] \times k \times (k+1) = -(k-1)k(k+1) + k(k+1)(k+2)\)

So we have

\(3S = 3 \times 1 \times 2 + 3 \times 2 \times 3 + {...} + 3 \times (n-1) \times n\)
\( = - (0 \times 1 \times 2 )+ (1 \times 2 \times 3) - (1 \times 2 \times 3) + (2 \times 3 \times 4) - {...} - [(n-2)(n-1)n] + [(n-1)n(n+1)]\)
\(= (n-1)n(n+1)\)

Hence \(S= \frac{(n-1)n(n+1)}{3}\)

Thus \(K = \frac{21 \times 22 \times 23}{3} = 7 \times 22 \times 23\). Answer E
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Now, go back to the formula of the sum of square consecutive. How we could find that formula? Here is it.

\(M = 1^2 + 2^2 + {...} + n^2\)

Note that \(k^2 = k \times k = k \times (k-1 + 1) = k(k-1) + k\), we have

\(M = 1^2 + 2^2 + {...} + n^2 = (0 \times 1 + 1) + (1 \times 2 + 2) + {...} + (n(n-1) + n)\)
\(= (1 \times 2 + 2 \times 3 + {...} + (n-1) \times n) + (1 + 2+ {...} + n)\)
\(= M1 + M2\)

We could easily have \(M2 = 1 + 2+ {...} + n = \frac{n(n+1)}{2}\)

Now, we have to calculate M1.

From my previous post, I have
\(M1 = 1 \times 2 + 2 \times 3 + {...} + (n-1) \times n = \frac{(n-1)n(n+1)}{3}\)

Thus, we have

\(M = M1 + M2 = \frac{(n-1)n(n+1)}{3} +\frac{n(n+1)}{2} = \frac{n(n+1)(2n+1)}{6}\)
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