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1^2 + 2.2^2 + 3^2 + 2.4^2 + ... up to n Terms Equals 1575: Find n | JEE

JEE Maths question with a full step-by-step solution.

Question
If
12+2⋅22+32+2⋅42+52+2⋅62+⋯1^{2} + 2\cdot 2^{2} + 3^{2} + 2\cdot 4^{2} + 5^{2} + 2\cdot 6^{2} + \cdots
 up to nn terms is equal to 15751575, then find the value of nn.
Solution
Answer: 14
Step 1: Read the pattern. The terms in odd positions are plain squares 12,32,52,…1^{2}, 3^{2}, 5^{2}, \ldots and those in even positions are doubled squares 2⋅22,2⋅42,2⋅62,…2\cdot2^{2}, 2\cdot4^{2}, 2\cdot6^{2}, \ldots Step 2: Rewrite the sum in a way that avoids two separate series. Every term is at least k2k^{2}, and the even ones carry one extra k2k^{2}, so
Sn=(12+22+⋯+n2)⏟all terms once+(22+42+62+⋯ )⏟the extra copy, even k≤n.S_n = \underbrace{\left(1^{2}+2^{2}+\cdots+n^{2}\right)}_{\text{all terms once}} + \underbrace{\left(2^{2}+4^{2}+6^{2}+\cdots\right)}_{\text{the extra copy, even }k \le n} .
 Step 3: Take nn even (we check the odd case in Step 8). The even squares up to nn are 22,42,…,n22^{2}, 4^{2}, \ldots, n^{2}, that is n2\dfrac{n}{2} terms, and taking out the factor 44,
22+42+⋯+n2=4(12+22+⋯+(n2)2)=4⋅n2(n2+1)(n+1)6=n(n+1)(n+2)6.2^{2}+4^{2}+\cdots+n^{2} = 4\left(1^{2}+2^{2}+\cdots+\left(\tfrac{n}{2}\right)^{2}\right) = 4 \cdot \frac{\frac{n}{2}\left(\frac{n}{2}+1\right)(n+1)}{6} = \frac{n(n+1)(n+2)}{6} .
 Step 4: Add the two pieces, using ∑k=1nk2=n(n+1)(2n+1)6\sum_{k=1}^{n}k^{2} = \dfrac{n(n+1)(2n+1)}{6}.
Sn=n(n+1)(2n+1)6+n(n+1)(n+2)6=n(n+1)[(2n+1)+(n+2)]6.S_n = \frac{n(n+1)(2n+1)}{6} + \frac{n(n+1)(n+2)}{6} = \frac{n(n+1)\left[(2n+1)+(n+2)\right]}{6} .
 Step 5: Simplify the bracket.
Sn=n(n+1)(3n+3)6=3n(n+1)26=n(n+1)22.S_n = \frac{n(n+1)(3n+3)}{6} = \frac{3n(n+1)^{2}}{6} = \frac{n(n+1)^{2}}{2} .
 Step 6: Set this equal to 15751575.
n(n+1)22=1575  ⟹  n(n+1)2=3150.\frac{n(n+1)^{2}}{2} = 1575 \implies n(n+1)^{2} = 3150 .
 Step 7: Factorise 31503150 to spot the answer.
3150=2×32×52×7=14×225=14×152,3150 = 2 \times 3^{2} \times 5^{2} \times 7 = 14 \times 225 = 14 \times 15^{2} ,
 so n=14n = 14 - and it is even, as assumed. Step 8: Rule out an odd nn. The same argument with nn odd (the extra copy then runs over 22,42,…,(n−1)22^{2}, 4^{2}, \ldots, (n-1)^{2}) gives Sn=n2(n+1)2S_n = \dfrac{n^{2}(n+1)}{2}, so we would need n2(n+1)=3150n^{2}(n+1) = 3150. But 132×14=236613^{2}\times 14 = 2366 and 152×16=360015^{2}\times 16 = 3600, so no odd nn works.
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