\(\sqrt [ 3 ]{ x^3+7 } ={(x^3+7)}^{{1\over 3}}\)
\(={\left[ x^3\left( 1+{7\over x^3} \right) \right]}^{{1\over 3}}\)   (\(\left |{7\over x^3}\right |<1\) as x is large)
\(=x{\left( 1+{7\over x} \right)}^{1\over 3}\)
\(=x\left( 1+{1\over 3} \times {7\over x^3}+{{{1\over3}\left( {1\over 3}-1 \right)}\over{2!}} {\left( {{7\over x^3}} \right)}^{2} +......\right)\)
\(=x\left( 1+{7\over 3}\times{1\over x^3}-{49\over 9}\times{1\over x^6}+...... \right)\)
\(=x+{7\over 3}\times{1\over x^2}-{49\over 9}\times{1\over x^5}+...\)
\(\sqrt [ 3 ]{ x^3+4 } ={(x^3+4)}^{1\over 3}\)
\(={\left[ x^3\left( 1+{4\over x^3} \right) \right]}^{1\over 3}\)
\(=x\left( 1+{4\over x^3} \right)^{1\over 3}\)
\(=x{\left( 1+{1\over3}\times{4\over x^3}+{{1\over 3}\left( {1\over3}-1 \right)\over{2!}} {\left( {4\over x^3} \right)}^{2}+... \right)}^{1\over3}\)
\(=x+{4\over 3}\times{1\over x^3}-{16\over 9}\times{1\over x^5}+...\)
Since x is large, \({1 \over x}\) is very small and hence higher powers of \({1 \over x}\) are negligible.
Thus \(\sqrt [ 3 ]{ x^3+7 } =x+{7\over 3}\times{1\over x^2}\) and \(\sqrt [ 3 ]{ x^3+4 } =x+{4\over3}\times{1\over x^3}.\) Therefore
\(\sqrt [ 3 ]{ x^3+7 } -\sqrt [ 3 ]{x^3+4 } =\left(x+{7\over 3}\times{1\over x^2} \right)-\left( x+{4\over 3}\times{1\over x^2} \right)={1\over x^2}\)

We know that for |x| < 1
\((1+x)^n=1+nx+{n(n-1)\over 2!}x^2+{n(n-1)(n-2)\over3!}x^3+....\)
\(\sqrt [ 3 ]{ 65} ={65}^{{1\over3}}\)
\(={(64+1)}^{1\over 3}\)
\(={64}^{1\over 3}{\left( 1+{1\over 64} \right)}^{1\over3}\Rightarrow4{\left( 1+{1\over64} \right)}^{1\over 3}\)
\(=4\left( 1+{1\over 3}\times{1\over 64}+{{1\over 3}\left({1\over3 }-1\right)\over{2!}} \times {\left({1\over 64} \right)}^{2}+....... \right)\)
\(=4+{1\over 48}-4\times{1\over 9}\times{1\over 64}\times{1\over 64}+....\)
\(=4+{1\over 48}-{4\over 36864}+...\)
\(=4+{1\over48}-{1\over 9216}+....\)
\(\approx 4 + 0.02 \)      ( since \({1\over 9216}+...\) is very small )
\(\sqrt [ 3 ]{ 65 }=4.02\) (approximately)

Let t_k denote the k^th term of the given series.
Then \(t_k={1\over \sqrt{k}+\sqrt{k+1}}.\) 
If we are successful in writing the k^th term as a difference of two expressions, then we can solve using this technique.
We have
\(t_k={1\over\sqrt{k}+\sqrt{k+1}}={{\sqrt{k}-\sqrt{k+1}}\over{(\sqrt{k}+\sqrt{k+1})(\sqrt{k}-)\sqrt{k+1}}}={{\sqrt{k}-\sqrt{k+1}}\over{k-(k+1)}}=\sqrt{k+1}-\sqrt{k}\)
Thus, \(t_1+t_2+....+t_4=(\sqrt{2}-\sqrt{1})+(\sqrt{3}-\sqrt{2})+...+(\sqrt{n-1}-\sqrt{n})=\sqrt{n+1}-1\)

We have 7⁴⁰⁰ = (7²)²⁰⁰ = (50 - 1)²⁰⁰
= ²⁰⁰C₀ 50²⁰⁰ - ²⁰⁰C₁ 50¹⁹⁹ + .... + ²⁰⁰C₁₉₈ 50²(-1)¹⁹⁸ + ²⁰⁰C₁₉₉ 50(-1)¹⁹⁹ + ²⁰⁰C₂₀₀(-1)²⁰⁰
= 50²-(²⁰⁰C₀50¹⁹⁸ - ²⁰⁰C₁50¹⁹⁷ + ..... + ²⁰⁰C₁₉₈(-1)¹⁹⁸) - 200 \(\times\) 50 + 1
As 50² and 200 are divisibie By 100, the last two digits: 0, 1.

To prove this it is enough to prove, 6ⁿ - 5n = 25k + 1 for some integer k. We first consider the expansion
(1+ x )ⁿ = ⁿC₀ + ⁿC₁ x + ⁿC₂ x² + ... + ⁿC_n-1 x^n-1 + ⁿC_n xⁿ, \(n \in N\)
Taking x = 5 we get (1 + 5)ⁿ = ⁿC₀ + ⁿC₁ 5 + ⁿC₂ 5² + ... + ⁿC_n-1 5^n-1 + ⁿC_n5n.
The above equality reduces to 6ⁿ = 1 + 5n + 25 (ⁿC₂ + 5ⁿC₃ + ... + ⁿC_n 5^n-2).
That is, 6ⁿ - 5n = 1 + 25(ⁿC₂ + 5 ⁿC₃ + ... + ⁿC_n 5^n-2) = 1 + 25k, \(k\in N\)
Thus 6ⁿ - 5n always leaves remainder 1 when divided by 25 for all positive integer n.

It is given that T₂ = 240, T₃ =·720 and T₄ = 1080.
T₂ = ⁿC₁ x^n-1 = 240 ...(1)
T₃ = ⁿC₂ x^n-2a² = 720 ....(2)
T₄ = ⁿC₃ x^n-3 a³ = 1080 ....(3)
Dividing (2) by (1) and (3) by (2) we get
\({a\over x}={6\over n-1}\)  ...(4)
\({a \over x}={9\over2(n-2)}\) .....(5)
From (4) and (5)    \({6\over n-1}={9\over 2(n-2)}\)
Thus n = 5. Substituting n = 5 in (1), (4) and dividing (1) by (4)
\({{5x^4 a}\over{{{a}\over{x}}}}={{240}\over{{6}\over{4}}}\)
Thus 5x⁵ = 160 and hence x = 2. Substituting in (4) we get a = 3.

\(\frac { 3-4x+{ x }^{ 2 } }{ { e }^{ 2x } } =\left( 3-4x+{ x }_{ 2 } \right) { e }^{ -2x }\)
\(=\left( 3-4x+{ x }^{ 2 } \right) \left( 1-\frac { 2x }{ 1! } +\frac { { \left( 2x \right) }^{ 2 } }{ 2! } -\frac { { \left( 2x \right) }^{ 3 } }{ 3! } +\frac { { \left( 2x \right) }^{ 4 } }{ 4! } -... \right) \)
\(=\left( 3-4x+{ x }^{ 2 } \right) \left( 1-2x+\frac { { 4x }^{ 2 } }{ 2 } -\frac { { 8 }^{ 3 } }{ 6 } +\frac { { 16 }^{ 4 } }{ 24 } -... \right) \)
\(=\left( 3-4x+{ x }^{ 2 } \right) \left( 1-2x+{ 2x }^{ 3 }-\frac { { 4x }^{ 3 } }{ 3 } +\frac { { 2x }^{ 4 } }{ 3 } -... \right) \)
\(=2+\frac { 16 }{ 3 } +2=\frac { 6+16+6 }{ 3 } =\frac { 28 }{ 3 } \)
Hence coefficient of x⁴ is \(\frac { 28 }{ 3 } \)

Let p = q+h
h is numerically very small and so h₁² h³ ... may be neglected
RHS \(={(n+1)p+(n-1)q\over (n-1)p+(n+1)q}={(n+1)(q+h)+(n-1)q\over (n+1)(q+h)+(n+1)q}\)
\(={nq+q-nh+h+nq-q\over nq-q+nh-h+nq+q}={2nq+(n+1)h\over2nq+(n-1)h}\)
\(={1+{n+1\over2n}.{q\over h}\over1+{n-1\over2n}.{h\over q}}=\left(1+{n+1\over2n}.{h\over q}\right)\left(1+{n-1\over2n}.{h\over q}\right)^{-1}\)
\(=\left({1+{n+1\over2n}.{h\over q}}\right)\left(1-{n-1\over2n}.{h\over q}\right)=1+\left({n+1\over 2n}-{n-1\over2n}\right){h\over q}\)
\(={1+{1\over n}}.{h\over q}\)
LHS \(={p\over q}^{1\over n}=\left(q+h\over q\right)^{1\over n}=\left(1+{h\over q}\right)^{1\over n}=1+{1\over n}.{h\over q}\)
From (1) and (2), LHS = RHS
Now \(\sqrt[8]{15\over16}={(8+1)(15)+(8-1)(16)\over(8-1)(15)+(8+1)(16)}\)  [n = 8, p = 15 and q = 16]
 \(={(9)(15)+7(16)\over(7)(15)+9(16)}={135+112\over105+144}={247\over 249}\)
\(\left( \frac { 15 }{ 16 } \right) =0.9919\)

In (1 + x)ⁿ, the general term is
T_r+1 = nCr (1)^n-r x^r ....(1)
To find the co-efficients of 5^th... , 6^th .. and 7^th ... terms, put r = 4, 5, 6 in (1) respectively.
∴Coefficients of 5^th ... , 6^th .... and 7^th ... terms in (1 +x)ⁿ are nC₄, nC₅ and nC₆ respectively.
Given that nC₄, nC₅ and nC₆ are in A.P
\(∴\ 2nC_5=nC_4+nC_6\)
\(⇒\ 2={nC_4\over nC_5}+{nC_6\over nC_5}\)    \(\left[∵{nC_r\over nC_{r-1}}={n-r+1\over r}\right]\)
\(⇒ 2={5\over n-4}+{n-5\over 6}\) 
\(⇒\ 12={30+(n-5)(n-4)\over 6(n-4)}\)

⇒ 12(n - 4) = 30 + n² - 4n - 5n + 20
⇒ 12n - 48 = n² - 9n + 50
⇒ n² - 9n + 50 - 12n + 48 = 0
⇒ n² - 21n + 98 = 0
⇒ (n - 14)(n - 7) = 0
⇒ n = 7 or 14.

Let the three consecutive terms be r^th...  (r + 1)^th ... and (r + 2)^th... terms.
General term in (x + a)ⁿ is T_{r+ 1} = nC_rx^n-r a^r     ... (1)
Then, co-efficients of r^th..  (r + 1)^th .. and (r + 2)^th .. terms are nC_r-1, nC_r and nC_r+1 respectively.
Given that nC_r-1 : nC_r: nC_r+1 = 1:7:42
Consider \({nC_{r-1}\over nC_r}={1\over7}\)
\({{n!\over (r-1)!(n-r+1)!}\over {n!\over r!(n-r)!}}={1\over 7}\)    \(\left[ ∵ nC_r \right]={n!\over r!(n-r)!}\)
\({n!\over(r-1)!(n-r+1)(n-r)!}\times{r(r-1)!(n-r)!\over n1}={1\over 7}\)
\({r\over n-r+1}={1\over 7}\)
⇒ 7r = n-r+1
⇒ n - 8r+ 1 = 0     ....(2)
and \({nC_r\over nC_{r+1}}r+1{}={7\over 42}\)
\({r+1\over n-r}={1\over 6}\)   \(\left[ ∴{nC_{r+1}\over nC_r}={n-r\over r+1}\right]\)
⇒ 6r + 6 = n-r
⇒ n -7r- 6 = 0 ....(3)
(2) - (3)
(n - 8r+ 1) - (n -7r- 6) = 0
⇒ -r + 7 = 0
⇒ r = 7
Substituting r = 7 in (2) we get
n - 8(7) + 1 = 0
⇒ n - 56 + 1 = 0
⇒ n-55 = 0
⇒ n = 55
Hence n = 55 and r = 7.