JEE Main 2023 — Mathematics Algebra

Given,

$$S={1}^{2}-2\cdot {3}^{2}+3\cdot {5}^{2}............+15\cdot {29}^{2}$$S=12−2⋅32+3⋅52............+15⋅292

Now rewriting the above expression we get,

$$S=\underset{{S}_{1}}{\underset{⏟}{{1}^{2}+2\cdot {3}^{2}+3\cdot {5}^{2}............+15\cdot {29}^{2}}}-2\underset{{S}_{2}}{\underset{⏟}{[2\cdot {3}^{2}+4\cdot {7}^{2}............+14\cdot {27}^{2}]}}$$S=S1​⏟12+2⋅32+3⋅52............+15⋅292​​−2S2​⏟[2⋅32+4⋅72............+14⋅272]​​

Now solving ${S}_{1}={1}^{2}+2\cdot {3}^{2}+3\cdot {5}^{2}............+15\cdot {29}^{2}$S1​=12+2⋅32+3⋅52............+15⋅292

$$\Rightarrow {S}_{1}=\sum _{r=1}^{15}r\cdot {(2r-1)}^{2}$$⇒S1​=r=1∑15​r⋅(2r−1)2

$$\Rightarrow {S}_{1}=\sum _{r=1}^{15}r\cdot (4{r}^{2}+1-4r)$$⇒S1​=r=1∑15​r⋅(4r2+1−4r)

$$\Rightarrow {S}_{1}=\sum _{r=1}^{15}(4{r}^{3}+r-4{r}^{2})$$⇒S1​=r=1∑15​(4r3+r−4r2)

$$\Rightarrow {S}_{1}=(4\sum _{r=1}^{15}{r}^{3}+\sum _{r=1}^{15}r-4\sum _{r=1}^{15}{r}^{2})$$⇒S1​=(4r=1∑15​r3+r=1∑15​r−4r=1∑15​r2)

$$\Rightarrow {S}_{1}=(4{(\frac{15\times 16}{2})}^{2}+\frac{15\times 16}{2}-4\frac{15\times 16\times 31}{6})$$⇒S1​=(4(215×16​)2+215×16​−4615×16×31​)

$$\Rightarrow {S}_{1}=(4{(120)}^{2}+120-4960)$$⇒S1​=(4(120)2+120−4960)

$$\Rightarrow {S}_{1}=52760$$⇒S1​=52760

Now solving ${S}_{2}=2\cdot {3}^{2}+4\cdot {7}^{2}............+14\cdot {27}^{2}$S2​=2⋅32+4⋅72............+14⋅272

$$\Rightarrow {S}_{2}=\sum _{r=1}^{7}(2r){(4r-1)}^{2}$$⇒S2​=r=1∑7​(2r)(4r−1)2

$$\Rightarrow {S}_{2}=\sum _{r=1}^{7}(2r)(16{r}^{2}+1-8r)$$⇒S2​=r=1∑7​(2r)(16r2+1−8r)

$$\Rightarrow {S}_{2}=32\sum _{r=1}^{7}{r}^{3}+2\sum _{r=1}^{7}r-16\sum _{r=1}^{7}{r}^{2}$$⇒S2​=32r=1∑7​r3+2r=1∑7​r−16r=1∑7​r2

$$\Rightarrow {S}_{2}=32{(\frac{7\times 8}{2})}^{2}+2(\frac{7\times 8}{2})-16(\frac{7\times 8\times 15}{6})$$⇒S2​=32(27×8​)2+2(27×8​)−16(67×8×15​)

$$\Rightarrow {S}_{2}=32\times 784+56-16\times 140=22904$$⇒S2​=32×784+56−16×140=22904

Hence, $S={S}_{1}-2{S}_{2}=52760-2\times 22904=6952$S=S1​−2S2​=52760−2×22904=6952